On the exactness of estimates for irregularly structured bodies of the general term of Laplace series

Abstract

The main form of the representation of a gravitational potential V for a celestial body T in outer space is the Laplace series in solid spherical harmonics $(R/r)^{n+1}Y_n(\theta ,\lambda )$ with R being the radius of the enveloping T sphere. The surface harmonic $Y_n$ satisfies the inequality

$$\begin{aligned} \langle Y_n\rangle < Cn^{-\sigma }. \end{aligned}$$

The angular brackets mark the maximum of a function’s modulus over a unit sphere. For bodies with an irregular structure $\sigma = 5/2$, and this value cannot be increased generally. However, a class of irregular bodies (smooth bodies with peaked mountains) has been found recently in which $\sigma = 3$. In this paper, we will prove the exactness of this estimate, showing that a body belonging to the above class does exist and

$$\begin{aligned} 0<\varlimsup n^3\langle Y_n\rangle <\infty \end{aligned}$$

for it.

Appendices

Appendix 1: Connection between Maclaurin coefficients of two functions

• 1a. Let

  $$\begin{aligned} f(z)= & {} \sum _{n=0}^{\infty } a_nz^{n+1},\qquad a_n=\frac{A_n}{(n+1)^\sigma }\gamma ^n, \qquad |A_n|\leqslant A, \nonumber \\ g(z)= & {} \frac{f(z)}{1-\beta z}=\sum _{n=0}^{\infty } b_nz^{n+1},\qquad b_n=\sum _{k=0}^{n} a_k\beta ^{n-k} :=\frac{B_n}{(n+1)^\sigma }\gamma ^n, \nonumber \\ \sigma\in & {} {\mathbb {R}},\quad 0<\beta <\gamma . \end{aligned}$$

  (47)

  Then

  $$\begin{aligned} -\infty<A_*=\varliminf A_n\leqslant \varlimsup A_n=A^*< \infty , \end{aligned}$$

  (48)

  the sequence $B_n$ is bounded

  $$\begin{aligned} |B_n|<\frac{A}{1-\delta }\times \left\{ \begin{array}{ll} 1,&{}\quad \text {if}\quad \sigma \leqslant 0,\\ 2^\sigma +\frac{1-\delta }{2\sqrt{\delta }} \left[ \frac{2(\sigma +1)}{e\ln (1/\delta )}\right] ^{\sigma +1}, &{}\quad \text {if}\quad \sigma > 0, \end{array}\right. \end{aligned}$$

  (49)

  and relations

  $$\begin{aligned} B_*=\varliminf B_n\geqslant \frac{A_*}{1-\delta }\,,\quad B^*=\varlimsup B_n\leqslant \frac{A^*}{1-\delta } \quad \text {with} \quad \delta =\frac{\beta }{\gamma }<1 \end{aligned}$$

  (50)

  are fulfilled. In particular, if the limit $\lim A_n=A^*$ exists, then the limit $\lim B_n=B^*$ exists also, and

  $$\begin{aligned} B^*=\frac{A^*}{1-\delta }. \end{aligned}$$

  (51)

  Proof Relations (48) represent a trivial corollary of the boundedness of $A_n$. It follows from (47) that

  $$\begin{aligned} B_n=\sum _{k=0}^{n} \left( \frac{n+1}{k+1}\right) ^\sigma \delta ^{n-k}A_k\,, \qquad \frac{|B_n|}{A}\leqslant \sum _{k=0}^{n} \left( \frac{n+1}{k+1}\right) ^\sigma \delta ^{n-k}:=u_n. \end{aligned}$$

  (52)

  If $\sigma \leqslant 0$, then $|B_n|< A(1-\delta )^{-1}$. Let $\sigma >0$. The last sum can be decomposed as

  $$\begin{aligned} u_n=u_{1n}+u_{2n} \end{aligned}$$

  with

  $$\begin{aligned}&u_{1n}:=\sum _{k=0}^{\lfloor n/2 \rfloor }\left( \frac{n+1}{k+1}\right) ^\sigma \delta ^{n-k}< (n+1)^\sigma \delta ^{\lceil n/2 \rceil } u_{3n}\,,\qquad u_{3n}=\sum _{k=0}^{\lfloor n/2 \rfloor } \frac{1}{(k+1)^\sigma }\,; \\&u_{2n}:= \sum _{k=\lfloor n/2 \rfloor +1}^n\left( \frac{n+1}{k+1}\right) ^\sigma \delta ^{n-k}< 2^\sigma \sum _{m=0}^{\infty }\delta ^{m}=\frac{2^\sigma }{1-\delta }. \end{aligned}$$

  If n is even and positive, then

  $$\begin{aligned} u_{3n}=1+\frac{1}{2^\sigma }+\cdots +\frac{1}{(1+n/2)^\sigma }<\frac{n+1}{2}. \end{aligned}$$

  It can be shown by induction. If n is odd, then the number of terms in the sum equals to $(n+1)/2$, and we arrive to the same inequality. Hence,

  $$\begin{aligned} 2u_{1n}<(n+1)^{\sigma +1} \delta ^{n/2}. \end{aligned}$$

  The right hand side has a maximum at

  $$\begin{aligned} n=\frac{2(\sigma +1)}{\ln (1/\delta )}-1. \end{aligned}$$

  So

  $$\begin{aligned} 2u_{1n}<\frac{1}{\sqrt{\delta }}\left[ \frac{2(\sigma +1)}{e\ln (1/\delta )}\right] ^{\sigma +1}. \end{aligned}$$

  The validity of (49) is proved for $n>0$. At $n=0$ the inequality (49) is trivial. Equalities

  $$\begin{aligned} b_{n+1}=a_{n+1}+\beta b_{n}\,,\qquad B_{n+1}=A_{n+1}+\delta \left( \frac{n+2}{n+1}\right) ^\sigma B_n \end{aligned}$$

  (53)

  follow from (47). The second formula (53) implies (Nikolsky 1977, section 3.7) that

  $$\begin{aligned} B_*\geqslant A_* +\delta B_*\,,\qquad B^*\leqslant A^* +\delta B^*. \end{aligned}$$

  (54)

  Relations (50) arise from (54). In case $A_*=A^*$ the equality (51) follows from (50).

• 1b. Let relations (47) be true except the last one, and now $\beta <0$, $|\beta |<\gamma$.

  Then inequalities (49) remain valid after replacing $\delta$ by $|\delta |$. Relations (53) are valid too. However, we get

  $$\begin{aligned} B_*\geqslant \frac{A_*-|\delta | A^*}{1-\delta ^2}\,,\qquad B^*\leqslant \frac{A^*-|\delta | A_*}{1-\delta ^2} \end{aligned}$$

  (55)

  instead of (50). If the limit $\lim A_n=A^*$ exists, then the limit $\lim B_n=B^*$ exists also, and the equality (51) holds true.

  Inequality (55) only needs a proof. Negativeness of $\delta$ implies $\varliminf \{\delta B_n\}=\delta \varlimsup B_n$, $\varlimsup \{\delta B_n\}=\delta \varliminf B_n$, see (Nikolsky 1977, section 3.7). We have now

  $$\begin{aligned} B_*\geqslant A_* -|\delta | B^*\,,\qquad B^*\leqslant A^* -|\delta | B_*\,, \end{aligned}$$

  (56)

  instead of (54). Inequalities (55) arise from (56).

• 1c. Let f(z) be holomorphic in the circle $|z|<1/|\beta |$, $\beta \ne 0$, and the series

  $$\begin{aligned} f(z)=\sum _{n=0}^{\infty } a_n\beta ^nz^n \end{aligned}$$

  converge (perhaps conditionally) at $z=1/\beta$.

  Then Maclaurin coefficients of the function

  $$\begin{aligned} g(z)=\frac{f(1/\beta )-f(z)}{1-\beta z}=\sum _{n=0}^{\infty } b_n\beta ^nz^n \end{aligned}$$

  are equal to

  $$\begin{aligned} b_n=f(1/\beta )-\sum _{m=0}^{n} a_m=\sum _{m=n+1}^{\infty } a_m. \end{aligned}$$

  (57)

  Proof Evidently,

  $$\begin{aligned} g(z)=\sum _{k=0}^{\infty }\beta ^kz^k\left[ f(1/\beta )- \sum _{m=0}^{\infty }a_m\beta ^mz^m\right] \end{aligned}$$

  in the circle $|z|<1/|\beta |$. The left equality (57) follows from it. The right equality (57) arises from the left one taking into account the convergence of the last series.

• 1d. Let

  $$\begin{aligned} f(z)= & {} \sum _{n=0}^{\infty } a_nz^{n+1},\qquad a_n=A_n\gamma ^n, \end{aligned}$$

  (58)
  $$\begin{aligned} g(z)= & {} \frac{f(z)}{1-\beta z}=\sum _{n=0}^{\infty } b_nz^{n+1},\qquad b_n=\sum _{k=0}^{n} a_k\beta ^{n-k}. \end{aligned}$$

  (59)

  Suppose

  $$\begin{aligned} \gamma>0,\qquad |\beta |>\gamma ,\qquad |A_n|\leqslant A, \qquad A>0. \end{aligned}$$

  Then

  $$\begin{aligned} b_n=B_n\beta ^n,\qquad B_n=\sum _{k=0}^{n} A_k\left( \frac{\gamma }{\beta }\right) ^{k}, \qquad |B_n|<B=\frac{A}{1-\gamma /|\beta |}. \end{aligned}$$

  (60)

  We omit an elementary proof.

Appendix 2: Representation of a certain standard function

The formula

$$\begin{aligned} \sqrt{1-2xz+z^2}=1-xz-\sum _{n=1}^{\infty }P_{n1}(x)z^{n+1} \end{aligned}$$

(61)

is valid (Antonov et al. 2010; Kholshevnikov and Shaidulin 2014) on the product of a segment $-1\leqslant x\leqslant 1$ and a circle $|z|\leqslant 1$. Here

$$\begin{aligned} P_{n1}(x)=\int _{-1}^xP_{n}(y)\,dy, \end{aligned}$$

$P_{n}$ being Legendre polynomial with the standard normalization $P_{n}(1)=1$.

Functions $P_{n1}$ have the following properties (Kholshevnikov and Shaidulin 2014, 2015b):

$$\begin{aligned} P_{n1}(\cos \theta )= & {} \sqrt{\frac{2\sin \theta }{\pi n^3}}\left\{ \cos \left[ \left( n+\frac{1}{2}\right) \theta +\frac{\pi }{4}\right] +\frac{r(n,\theta )}{n\sin \theta }\right\} ,\nonumber \\ |P_{n1}(\cos \theta )|< & {} \frac{\sqrt{2/\pi }}{n^{3/2}}. \end{aligned}$$

(62)

Here $r(n,\theta )$ is bounded under $n\geqslant 1$, $0\leqslant \theta \leqslant \pi$; the exponent 3 / 2 in the estimate of $|P_{n1}|$ is exact. Moreover, $P_{n1}(0)=0$ if n is even, and

$$\begin{aligned} P_{n1}(0)=(-1)^{(n+1)/2}\frac{(n-2)!!}{(n+1)!!}\sim (-1)^{(n+1)/2} \frac{\sqrt{2/\pi }}{n^{3/2}} \end{aligned}$$

(63)

if n is odd.

Appendix 3: Asymptotics of a certain integral

An asymptotic representation (under $\nu \rightarrow \infty$) with a remainder

$$\begin{aligned} \int y^\nu \,dx=\sum _{n=0}^{k}c_n\frac{y^{\nu +n+1}}{(x-bc)^{2n+1}}+(2k+1)c_k\int \frac{y^{\nu +k+1}\,dx}{(x-bc)^{2k+2}} \end{aligned}$$

(64)

is true. Here

$$\begin{aligned} y=x^2-2bcx+b^2,\qquad c_n=\frac{(2n-1)!!}{2^{n+1}(\nu +1)^{\overline{n+1}}}, \end{aligned}$$

the point $x_0=bc$ must not belong to the segment of integration.

To prove (64) it is sufficient to differentiate it.

Limiting ourselves to the first two terms in the right hand side of (64) we obtain

$$\begin{aligned} \int y^\nu \,dx\asymp \frac{y^{\nu +1}}{2(\nu +1)(x-bc)}+ \frac{y^{\nu +2}}{4(\nu +1)(\nu +2)(x-bc)^3}+\cdots \end{aligned}$$

(65)

As a consequence of (64) it is easy to establish, that

$$\begin{aligned} \int xy^\nu \,dx= & {} \frac{xy^{\nu +1}}{2(\nu +1)(x-bc)}+bc\sum _{n=1}^{k}c_n \frac{y^{\nu +n+1}}{(x-bc)^{2n+1}}+ \nonumber \\&+(2k+1)bcc_k\int \frac{y^{\nu +k+1}\,dx}{(x-bc)^{2k+2}}. \end{aligned}$$

(66)

Limiting ourselves to the first two terms in the right hand side we receive

$$\begin{aligned} \int xy^\nu \,dx\asymp \frac{xy^{\nu +1}}{2(\nu +1)(x-bc)}+ \frac{bc y^{\nu +2}}{4(\nu +1)(\nu +2)(x-bc)^3}+\cdots \end{aligned}$$

(67)

Let I be the integral (67) taken between 0 and a, $a>0$. We assume the possibility that $x_0\in [0,a]$. Let us prove the asymptotic representation:

$$\begin{aligned} I=\left\{ \begin{array}{ll} \frac{y^{\nu +2}(0)}{4(\nu +1)(\nu +2)(bc)^2}+\cdots ,&{}\quad \text {if}\quad a-2bc<0,\\ \frac{y^{\nu +1}(a)}{\nu +1}+\frac{2y^{\nu +2}(a)}{(\nu +1)(\nu +2)a^2}+\cdots ,&{}\quad \text {if}\quad a-2bc=0,\\ \frac{ay^{\nu +1}(a)}{2(\nu +1)(a-bc)}+\frac{bc y^{\nu +2}(a)}{4(\nu +1)(\nu +2)(a-bc)^3}+\cdots ,&{}\quad \text {if}\quad a-2bc>0. \end{array}\right. \end{aligned}$$

(68)

The variable y(x) is a downward-convex function, and takes the maximum at one of the endpoints of the segment [0, a]. We consider three cases depending on the sign of the difference $y(a)-y(0)=a(a-2bc)$.

• (a) $a-2bc<0$, $\max y(x)=y(0)>y(a)$, $bc>a/2$. If $bc>a$, then we may use (67) straightforwardly. If $a/2<bc\leqslant a$ we may restrict ourselves to an integration over the segment [0, a / 2]. Indeed,

  $$\begin{aligned}&\int _{0}^{a/2} xy^\nu \,dx\sim \frac{bc y^{\nu +2}(0)}{4(\nu +1)(\nu +2)(bc)^3}\,, \\&\int _{a/2}^{a} xy^\nu \,dx<\frac{a^2}{2}{{\bar{y}}}^\nu ,\quad \bar{y}=\max \{y(a/2),\,y(a)\}<y(0). \end{aligned}$$

• (b) $a-2bc=0$, $\max y(x)=y(a)=y(0)$. The function y(x) is symmetric with respect to the point $x_0=bc=a/2$. After the substitution $x=x_0+z$ we have

  $$\begin{aligned} I=\frac{a}{2}\int _{-a/2}^{a/2} y^\nu \,dz+\int _{-a/2}^{a/2} zy^\nu \,dz= a\int _{0}^{a/2} y^\nu \,dz=a\int _{a/2}^{a} y^\nu \,dx. \end{aligned}$$

  The last integral over the segment from a / 2 to 3a / 4 may be thrown off. For the integral from 3a / 4 to a we may use (65).

• (c) $a-2bc>0$, $\max y(x)=y(a)>y(0)$. If $bc<0$, then $x_0$ lies out of the segment of integration, and (68) follows from (67). Let $bc\geqslant 0$. Then $bc<a/2$, $a-bc>a/2$, so

  $$\begin{aligned} \int _{0}^{a/2} xy^\nu \,dx<\frac{a^2}{8}{{\bar{y}}}^\nu ,\quad \bar{y}=\max \{y(0),\,y(a/2)\}<y(a). \end{aligned}$$

  For the integral between a / 2 and a we may use the formula (67).