Geometric Designs and Rotatable Designs I

Abstract

Victoir (2004) developes a powerful method for constructing high-dimensional cubature formulas with small number of points, which consists of the corner-vector approach for the hyperoctahedral group $B_d$ and the thinning approach based on orthogonal arrays and combinatorial designs. Hirao et al. (2014) applies this to constructions of optimal rotatable designs on the d-dimensional unit ball ${\varvec{B}}^d$ and thereby finds some infinite series of third-order designs with small number of points. Afterwards, Sawa and Hirao (2017) discusses the corner-vector approach for group $D_d$ that can be realized as the symmetry group of a demihypercube in ${\mathbb {R}}^d$. In the present paper we discuss a further generalization and improvement of these works, and describe many examples and infinite series of D-optimal rotatable designs with small number of points. The core and novelty in the presentation lies in the focus on the cubature theory in numerical analysis or Euclidean design theory in combinatorics, which not only improves and/or generalizes previous works in those fields of mathematics but also provides a unified mathematical description of various classes of factorial designs, including Box–Hunter solid designs, Box–Behnken designs, central composite designs and Plackett–Burman designs, in the modern framework of the cubature theory. Our new framework enables one, for example, to briefly evaluate the maximum integer t for which such classical factorial designs are of t-th order rotatable.

Appendices

A. Optimal Values

This paper deals with D-optimal designs for the standard e-th polynomial regression model

$$\begin{aligned} Y(\varvec{\omega }) = \sum _{\varvec{\alpha }} \theta _{\varvec{\alpha }} f_{\varvec{\alpha }} (\varvec{\omega }) + \epsilon (\varvec{\omega }), \qquad \varvec{\omega }\in {\mathbf {B}}^d, \end{aligned}$$

(8)

where each $f_{\varvec{\alpha }}$ is taken to be monomial $\varvec{\omega }^{\varvec{\alpha }}$, $\theta _{\alpha }$ are unknown parameters, and $\epsilon$ is a random variable.

Given a (possibly infinite) design $\xi$,

$$\begin{aligned} \mathbb {M}(\xi ) = \left( \int _{{\varvec{B}}^d} f_{\varvec{\alpha }} (\varvec{\omega }) f_{\varvec{\alpha }'} (\varvec{\omega }) \; \xi (d \varvec{\omega }) \right) _{\varvec{\alpha }. \varvec{\alpha }'} \end{aligned}$$

is the information matrix for model (8).

Let $X = \{ {\varvec{x}}_1, \ldots , {\varvec{x}}_n \}$ be observation points and $\xi := \xi _X = \frac{1}{n} \sum _{i} \delta _{{\varvec{x}}_i}$ be the corresponding design, where $\delta _{{\varvec{x}}}$ is the Dirac measure at point ${\varvec{x}}$. Assume that $\epsilon$ has a normal distribution with mean 0, variance $\sigma ^ 2 > 0$ and $E[\epsilon (\varvec{\omega }) \epsilon (\varvec{\omega }')] = \sigma ^2 \delta _{\varvec{\omega }, \varvec{\omega }'}$ for all $\varvec{\omega }, \varvec{\omega }' \in {\mathbf {B}}^d$. Let ${\hat{\varvec{\theta }}}$ be the least squares estimator of $\varvec{\theta }= (\theta _{\varvec{\alpha }})_{\varvec{\alpha }}$ in model (8). A confidence ellipsoid of ${\hat{\varvec{\theta }}}$ is then given as

$$\begin{aligned} \left\{ \varvec{\theta }\mid (\varvec{\theta }- {\hat{\varvec{\theta }}})^T \mathbb {M}(\xi ) (\varvec{\theta }- {\hat{\varvec{\theta }}}) \le \text {constant} \right\} ; \end{aligned}$$

see, e.g., [54]. We note that, the volume of ${\mathcal {X}}$ is proportional to $(\det \mathbb {M}(\xi ))^{-1/2}$. Thus D-optimality, which seeks $\xi$ maximizing $\det \mathbb {M}(\xi )$, is a popular and natural optimality criterion.

Since D-optimal weights and radii are determined without depending on the model; see Proposition 3.3 in [51], we treat the Zernike-type polynomial regression model (see [5, 42, 50]). We take $\{ f_{\alpha } \}_{\alpha }$ to be

$$\begin{aligned} {\mathcal {Q}}_{e}({\varvec{B}}^d) = \left\{ \Vert \varvec{\omega }\Vert ^{2j} \phi _{l,i} \mid 0 \le l \le e, \; 0 \le j \le \Big \lfloor \frac{e - l}{2} \Big \rfloor , \; 1 \le i \le h_l^d \right\} , \end{aligned}$$

where $h_l^d := \dim \mathop {\mathrm {Harm}}\nolimits _{l}({\mathbb {R}}^d) = \left( {\begin{array}{c}d + l - 1\\ l\end{array}}\right) - \left( {\begin{array}{c}d + l - 3\\ l -2\end{array}}\right)$ and $\phi _{l,i}$ is an i-th spherical harmonics of degree l.

A merit of Zernike-type polynomial regression model is to simplify calculations of $\det \mathbb {M}(\xi )$. For example when $e = 3$, the information matrix of $\xi$ is represented by

$$\begin{aligned} \det \mathbb {M}(\xi ) = \prod _{l = 0}^{3} (\det \mathbb {M}_{l}(\xi ))^{h_l^d}, \end{aligned}$$

where

$$\begin{aligned} \mathbb {M}_0 (\xi )&= \begin{pmatrix} 1 &{} W_1 + r_2^2 W_2 \\ W_1 + r_2^2 W_2 &{} W_1 + r_2^4 W_2 \\ \end{pmatrix}, \; \mathbb {M}_1 (\xi ) = \begin{pmatrix} W_1 + r_2^2 W_2 &{} W_1 + r_2^4 W_2 \\ W_1 + r_2^4 W_2 &{} W_1 + r_2^6 W_2 \\ \end{pmatrix}, \\ \mathbb {M}_2 (\xi )&= \begin{pmatrix} W_1 + r_2^4 W_2 \end{pmatrix}, \quad \mathbb {M}_3 (\xi ) = \begin{pmatrix} W_1 + r_2^6 W_2 \end{pmatrix}. \end{aligned}$$

In order to obtain the optimal values of $r_2$ and $W_2$, it suffices to solve the following equations:

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle \frac{d}{r_2^2} - \frac{2 (d + 1)}{1 - r_2^2} + \frac{2 h_2^d W_2 r_2^2}{1 - W_2 + W_2 r_2^4} + \frac{3 h_3^d W_2 r_2^4}{1 - W_2 + W_2 r_2^6} = 0, \\ \displaystyle \frac{(d + 1)(1 - 2 W_2)}{W_2 (1 - W_2)} - \frac{h_2^d (1 - r_2^4)}{1 - W_2 + W_2 r_2^4} - \frac{h_3^d (1 - r_2^6)}{1 - W_2 + W_2 r_2^6} = 0. \end{array} \right. \end{aligned}$$

(9)

Solving them numerically, the following optimal values are obtainable, e.g.,

• $d = 2^6$, $W_2 = 0.00143065$, $r_2^2 = 0.363407$;

• $d = 2^8$, $W_2 = 0.0000944923$, $r_2^2 = 0.365397$;

• $d = 2^{10}$, $W_2 = 5.98901 \times 10^{-6}$, $r_2^2 = 0.36587$.

Moreover, when $e = 2$, by the same procedure, the optimal values $W_i$ can be more easily computed as follows:

$$\begin{aligned} W_1 = \sum _{x \in X \setminus \{ {\varvec{0}} \}} w(x) = \frac{d(d+3)}{(d+1)(d+2)}, \quad W_2 = w({\varvec{0}}) = \frac{2}{(d+1)(d+2)}. \end{aligned}$$

(10)

Similarly for e in general, we can numerically compute the optimal weights $W_i$ and radii $r_i$ of a D-optima rotatable design of degree e on ${\varvec{B}}^d$. We omit the detail here, see, e.g., Sect.  3.3 in Sawa et al. [51].

B. Examples of Weighted Spherical Designs

We give two lists of $B_d$- and $D_d$-invariant weighted spherical 9- and 11-designs on $\mathbb {S}^{d-1}$ generated from the corner vector method. Since a huge number of examples are constructed, we only consider the situation where $a > 1$ with $a^2 \in \{ 2, \ldots , 8 \}$, $3 \le d \le 12$ and the number of orbits is at most 6, and only describe the smallest number of points in each dimension.

Table 2 weighted spherical 9-design

Table 3 weighted spherical 11-design

C. Invariant Harmonic Polynomials

In this section we describe informations of $D_d$-invariant harmonic polynomials that are used to construct (optimal) rotatable designs in Sects. 3, 4 and 5 .

1.1 C. 1 Group $B_d$

Harmonic Molien-Poincaré series

The harmonic Molien-Poincaré series is given by

$$\begin{aligned} \sum _{i=0}^{\infty } p_{i} \lambda ^{i} = \displaystyle \frac{1}{(1-\lambda ^4)(1-\lambda ^6) \cdots (1-\lambda ^{2d})} = \left\{ \begin{array}{ll} 1+\lambda ^4+\lambda ^6+\lambda ^8+\lambda ^{10}+ \cdots , &{} d=3, \\ 1+\lambda ^4+\lambda ^6+2\lambda ^8+\lambda ^{10}+ \cdots , &{} d=4, \\ 1+\lambda ^4+\lambda ^6+2\lambda ^8+2\lambda ^{10}+ \cdots , &{} d \ge 5. \end{array} \right. \end{aligned}$$

Dimensions of $B_d$-invariant harmonic polynomial spaces

By the above harmonic Molien–Poincaré series, dimensions of $B_d$-invariant harmonic polynomial spaces are given by

$$\begin{aligned} \dim \mathop {\mathrm {Harm}}\nolimits _{1}({\mathbb {R}}^d)^{B_{d}}&= \dim \mathop {\mathrm {Harm}}\nolimits _{2}({\mathbb {R}}^d)^{B_{d}} = 0, \quad d \ge 3, \\ \dim \mathop {\mathrm {Harm}}\nolimits _{3}({\mathbb {R}}^d)^{B_{d}}&= \dim \mathop {\mathrm {Harm}}\nolimits _{5}({\mathbb {R}}^d)^{B_d} = \dim \mathop {\mathrm {Harm}}\nolimits _{7}({\mathbb {R}}^d)^{B_d} = \dim \mathop {\mathrm {Harm}}\nolimits _{9}({\mathbb {R}}^d)^{B_d} = 0, \\&\qquad \qquad d \ge 3, \\ \dim \mathop {\mathrm {Harm}}\nolimits _{4}({\mathbb {R}}^d)^{B_d}&= \dim \mathop {\mathrm {Harm}}\nolimits _{6}({\mathbb {R}}^d)^{B_d} = 1, \quad d \ge 3, \\ \dim \mathop {\mathrm {Harm}}\nolimits _{8}({\mathbb {R}}^d)^{B_d}&= {\left\{ \begin{array}{ll} 1, & d = 3, \\ 2, & d \ge 4, \end{array}\right. } \\ \dim \mathop {\mathrm {Harm}}\nolimits _{10}({\mathbb {R}}^d)^{B_d}&= {\left\{ \begin{array}{ll} 1, & d d = 3,4, \\ 2, & d \ge 5. \end{array}\right. } \end{aligned}$$

Basis of $B_d$-invariant harmonic polynomial spaces By the direct calculations, it is easy to find a basis of $B_d$-invariant harmonic polynomial spaces $\mathop {\mathrm {Harm}}\nolimits _{\ell } ({\mathbb {R}}^d)^{B_d}$.

• The space $\mathop {\mathrm {Harm}}\nolimits _{4}({\mathbb {R}}^d)^{B_d}$ is spanned by polynomial $f_{4}$ defined as

  $$\begin{aligned} f_{4} = \text {sym}(x_{1}^{4}) - \frac{6}{d-1} \text {sym}(x_{1}^{2} x_{2}^2), \quad d \ge 3. \end{aligned}$$

• The space $\mathop {\mathrm {Harm}}\nolimits _{6}({\mathbb {R}}^d)^{B_d}$ is spanned by polynomial $f_{6}$ defined as

  $$\begin{aligned} f_{6} = \text {sym}(x_{1}^{6}) - \frac{15}{d-1} \text {sym}(x_{1}^{2} x_{2}^4) + \frac{180}{(d-1)(d-2)} \text {sym}(x_{1}^{2} x_{2}^{2} x_{3}^{2}), \quad d \ge 3. \end{aligned}$$

• The space $\mathop {\mathrm {Harm}}\nolimits _{8}({\mathbb {R}}^d)^{B_d}$ is spanned by polynomials $f_{8, 1}$ and $f_{8, 2}$ defined as

  $$\begin{aligned} f_{8,1}&= \text {sym}(x_{1}^{8}) - \frac{28}{d-1} \text {sym}(x_{1}^{2} x_{2}^{6}) + \frac{70}{d-1} \text {sym}(x_{1}^{4} x_{2}^{4}), \quad d \ge 3, \\ f_{8,2}&= \text {sym}(x_{1}^{4} x_{2}^{4}) - \frac{6}{d-2} \text {sym}(x_{1}^{2} x_{2}^{2} x_{3}^{4}) + \frac{108}{(d-2)(d-3)} \text {sym}(x_{1}^{2} x_{2}^{2}x_{3}^{2}x_{4}^{2}), \quad d \ge 4. \end{aligned}$$

• The space $\mathop {\mathrm {Harm}}\nolimits _{10}({\mathbb {R}}^d)^{B_d}$ is spanned by polynomials $f_{10, 1}$ and $f_{10, 2}$ defined as

  $$\begin{aligned} f_{10,1}&= \text {sym}(x_{1}^{10}) - \frac{45}{d-1} \text {sym}(x_{1}^{2} x_{2}^{8}) + \frac{42}{d-1} \text {sym}(x_{1}^{4} x_{2}^{6}) \\&\quad + \frac{1008}{(d-1)(d-2)} \text {sym}(x_{1}^{2} x_{2}^{2} x_{3}^{6}) \\&\quad - \frac{1260}{(d-1)(d-2)} \text {sym}(x_{1}^{2} x_{2}^{4} x_{3}^{4}),\quad d \ge 3, \\ f_{10,2}&= \text {sym}(x_{1}^{4} x_{2}^{6}) - \frac{6}{d-2} \text {sym}(x_{1}^{2} x_{2}^{2} x_{3}^{6}) - \frac{30}{d-2} \text {sym}(x_{1}^{2} x_{2}^{4} x_{3}^{4}) \\&\quad + \frac{450}{(d-2)(d-3)} \text {sym}(x_{1}^{2}x_{2}^{2}x_{3}^{2}x_{4}^{4}) \\&\quad - \frac{10800}{(d-2)(d-3)(d-4)} \text {sym}(x_{1}^{2} x_{2}^{2} x_{3}^{2} x_{4}^{2} x_{5}^{2}), \quad d \ge 5. \end{aligned}$$

Substituting ${\tilde{{\varvec{v}}}}_k$ and ${\tilde{{\varvec{v}}}}_{k, a}$ for $B_d$-invariant harmonic polynomials

• For $d \ge 3$

  $$\begin{aligned} f_{4}({\tilde{{\varvec{v}}}}_k)&= \frac{1}{k}\left( 1-\frac{3(k-1)}{d-1} \right) , \\ f_{4}({\tilde{{\varvec{v}}}}_{k, a})&= \frac{1}{(a^{2}+k)^{2}} \left( a^{4}+k - \frac{6}{d-1} \Bigl (a^{2}k + \frac{k(k-1)}{2} \Bigr )\right) . \end{aligned}$$

• For $d \ge 3$

  $$\begin{aligned} f_{6}({\tilde{{\varvec{v}}}}_k)&= \frac{1}{k^{2}} \left( 1-\frac{15(k-1)}{d-1} + \frac{30(k-1)(k-2)}{(d-1)(d-2)}\right) , \\ f_{6}({\tilde{{\varvec{v}}}}_{k,a})&= \frac{1}{(a^{2}+k)^{3}} \biggl ( a^{6}+k- \frac{15}{d-1} \left( a^{2}k + a^{4}k + k(k-1) \right) \\&\quad + \frac{180}{(d-1)(d-2)} \Bigl ( \frac{a^{2} k (k-1)}{2} + \frac{k(k-1)(k-2)}{3!} \Bigr )\biggr ). \end{aligned}$$

• For $d \ge 3$

  $$\begin{aligned} f_{8,1}({\tilde{{\varvec{v}}}}_k)&= \frac{1}{k^{3}} \left( 1+\frac{7(k-1)}{d-1} \right) , \\ f_{8,1}({\tilde{{\varvec{v}}}}_{k,a})&= \frac{1}{(a^{2}+k)^{4}} \biggl ( a^{8}+k- \frac{28}{n-1} \bigl (a^{2}k + a^{6}k +k(k-1) \bigr ) \\&\quad + \frac{70}{d-1} \Bigl ( a^{4}k + \frac{k(k-1)}{2}\Bigr )\biggr ). \end{aligned}$$

  For $d \ge 4$

  $$\begin{aligned} f_{8,2}({\tilde{{\varvec{v}}}}_k)&= \frac{k-1}{2k^{3}} \left( 1-\frac{6(k-2)}{d-2} + \frac{9(k-2)(k-3)}{(d-2)(d-3)} \right) , \\ f_{8,2}({\tilde{{\varvec{v}}}}_{k,a})&= \frac{1}{(a^{2}+k)^{4}} \biggl ( a^{4}+ \frac{k(k-1)}{2} - \frac{6}{d-2} \Bigl (a^{2}k(k-1) \\&\quad + \frac{a^{4}k(k-1)}{2} + \frac{k(k-1)(k-2)}{2} \Bigr ) \\&\quad + \frac{9k(4a^{2} + k-3)(k-1)(k-2)}{2(d-2)(d-3)}\biggr ). \end{aligned}$$

• For $d \ge 3$

  $$\begin{aligned} f_{10,1}({\tilde{{\varvec{v}}}}_k)&= \frac{1}{k^{4}} \left( 1- \frac{3(k-1)}{d-1} - \frac{126(k-1)(k-2)}{(d-1)(d-2)}\right) , \\ f_{10,1}({\tilde{{\varvec{v}}}}_{k,a})&= \frac{1}{(a^{2}+k)^{5}} \biggl ( a^{10}+k - \frac{3k(15a^{8} -14a^{6} -14a^{4} +15a^{2} +k-1)}{d-1} \\&\quad + \frac{126k(k-1)(4a^{6} -10a^{4} +3a^{2} -k+2)}{(d-1)(d-2)} \biggr ). \end{aligned}$$

  For $d \ge 5$

  $$\begin{aligned} f_{10,2}({\tilde{{\varvec{v}}}}_k)&= \frac{1}{k^{4}} \biggl ( k-1 - \frac{18(k-1)(k-2)}{d-2} + \frac{75(k-1)(k-2)(k-3)}{(d-2)(d-3)} \\&\quad - \frac{90(k-1)(k-2)(k-3)(k-4)}{(d-2)(d-3)(d-4)}\biggr ), \\ f_{10,2}({\tilde{{\varvec{v}}}}_{k,a})&= \frac{1}{(a^{2}+k)^{5}} \biggl ( a^{4}k+ a^{6}k + k(k-1) - \frac{3k(k-1)(a^{6}+10a^{4}+7a^{2}+6k-12)}{d-2} \\&\quad + \frac{75k(k-1)(k-2)(a^{4}+3a^{2}+k-3)}{(d-2)(d-3)} \\&\quad - \frac{90k(k-1)(k-2)(k-3)(5a^{2}+k-4)}{(d-2)(d-3)(d-4)} \biggr ). \end{aligned}$$

1.2 C. 2 Group $D_d$

Harmonic Molien-Poincaré series

The harmonic Molien–Poincaré series is given by

$$\begin{aligned} \sum _{j=0}^{\infty } q_{j} \lambda ^{j}&= \displaystyle \frac{1}{(1-\lambda ^4)(1-\lambda ^6) \cdots (1-\lambda ^{2n-2})(1-\lambda ^n)} \\&= \left\{ \begin{array}{ll} 1+2\lambda ^4+\lambda ^6+3\lambda ^8+2\lambda ^{10}+ \cdots , \, d=4, \\ 1+\lambda ^4+\lambda ^5+\lambda ^6+2\lambda ^8+\lambda ^9+2\lambda ^{10}+\lambda ^{11} \cdots , \, d=5, \\ 1+\lambda ^4+2\lambda ^6+2\lambda ^8+3\lambda ^{10}+ \cdots , \, d=6, \\ 1+\lambda ^4+\lambda ^6+\lambda ^7+2\lambda ^8+2\lambda ^{10}+\lambda ^{11} \cdots , \, d=7, \\ 1+\lambda ^4+\lambda ^6+3\lambda ^8+2\lambda ^{10}+ \cdots , \, d=8, \\ 1+\lambda ^4+\lambda ^6+2\lambda ^8+\lambda ^9+2\lambda ^{10}+ \cdots , \, d=9,\\ 1+\lambda ^4+\lambda ^6+2\lambda ^8+3\lambda ^{10}+ \cdots , \, d=10, \\ 1+\lambda ^4+\lambda ^6+2\lambda ^8+2\lambda ^{10}+\lambda ^{11} \cdots , \, d=11, \\ 1+\lambda ^4+\lambda ^6+2\lambda ^8+2\lambda ^{10}+ \cdots , \, d \ge 12 . \end{array} \right. \end{aligned}$$

Dimensions of $D_d$-invariant harmonic polynomial spaces

By the above harmonic Molien–Poincaré series, dimensions of $D_d$-invariant harmonic polynomial spaces are given by

$$\begin{aligned} \dim \mathop {\mathrm {Harm}}\nolimits _{1}({\mathbb {R}}^d)^{D_d}&= \dim \mathop {\mathrm {Harm}}\nolimits _{2}({\mathbb {R}}^d)^{D_d} = \dim \mathop {\mathrm {Harm}}\nolimits _{3}({\mathbb {R}}^d)^{D_d} = 0, \quad d \ge 4, \\ \dim \mathop {\mathrm {Harm}}\nolimits _{4}({\mathbb {R}}^d)^{D_d}&= {\left\{ \begin{array}{ll} 1, \quad d \ge 5, \\ 2, \quad d = 4, \end{array}\right. } \\ \dim \mathop {\mathrm {Harm}}\nolimits _{5}({\mathbb {R}}^d)^{D_d}&= {\left\{ \begin{array}{ll} 0, \quad d \ge 4, d \ne 5, \\ 1, \quad d= 5, \end{array}\right. } \\ \dim \mathop {\mathrm {Harm}}\nolimits _{6}({\mathbb {R}}^d)^{D_d}&= {\left\{ \begin{array}{ll} 1, \quad d \ge 4, d \ne 6, \\ 2, \quad d = 6, \end{array}\right. } \\ \dim \mathop {\mathrm {Harm}}\nolimits _{7}({\mathbb {R}}^d)^{D_d}&= {\left\{ \begin{array}{ll} 0, \quad d \ge 4, d \ne 7, \\ 1, \quad d = 7, \end{array}\right. } \\ \dim \mathop {\mathrm {Harm}}\nolimits _{8}({\mathbb {R}}^d)^{D_d}&= {\left\{ \begin{array}{ll} 2, \quad d \ge 5, d \ne 8, \\ 3, \quad d = 4,8, \end{array}\right. } \\ \dim \mathop {\mathrm {Harm}}\nolimits _{9}({\mathbb {R}}^d)^{D_d}&= {\left\{ \begin{array}{ll} 0, \quad d \ge 4, d \ne 5,9, \\ 1, \quad d = 5,9, \end{array}\right. } \\ \dim \mathop {\mathrm {Harm}}\nolimits _{10}({\mathbb {R}}^d)^{D_d}&= {\left\{ \begin{array}{ll} 2, \quad d \ge 4, d \ne 6,10, \\ 3, \quad d = 6,10, \end{array}\right. } \\ \dim \mathop {\mathrm {Harm}}\nolimits _{11}({\mathbb {R}}^d)^{D_d}&= {\left\{ \begin{array}{ll} 0, \quad d \ge 4, d \ne 5,7,11, \\ 1, \quad d = 5,7,11. \end{array}\right. } \end{aligned}$$

Basis of $D_d$-invariant harmonic polynomial spaces

By direct calculations, it is easy to find a basis of $D_d$-invariant harmonic polynomial spaces $\mathop {\mathrm {Harm}}\nolimits _{\ell }({\mathbb {R}}^d)^{D_d}$.

• The space $\mathop {\mathrm {Harm}}\nolimits _{4}({\mathbb {R}}^d)^{D_d}$ is spanned by polynomials $f_{4, 1}$ and $f_{4, 2}$ defined as

  $$\begin{aligned} f_{4,1}(x)&= \text {sym}(x_1^4) - \frac{6}{d-1} \text {sym}(x_1^2 x_2^2), \quad d \ge 5, \\ f_{4, 2}(x)&= x_1 x_2 x_3 x_4, \quad d = 4. \end{aligned}$$

• The space $\mathop {\mathrm {Harm}}\nolimits _{5}({\mathbb {R}}^5)^{D_5}$ is spanned by polynomial defined $f_{5}$ as

  $$\begin{aligned} f_5 = x_1 x_2 x_3 x_4 x_5. \end{aligned}$$

• $\mathop {\mathrm {Harm}}\nolimits _{6}({\mathbb {R}}^d)^{D_d}$ is spanned by polynomials $f_{6, 1}$ and $f_{6, 2}$ defined as

  $$\begin{aligned} f_{6, 1}&= \text {sym}(x_1^6) - \frac{15}{d-1} \text {sym}(x_1^2 x_2^4) + \frac{180}{(d - 1)(d - 2)} \text {sym} (x_1^2 x_2^2 x_3^2), \quad d \ge 4,\; d \ne 6, \\ f_{6, 2}&= x_1 x_2 x_3 x_4 x_5 x_6, \quad d = 6. \end{aligned}$$

• $\mathop {\mathrm {Harm}}\nolimits _{7}({\mathbb {R}}^7)^{D_7}$ is spanned by polynomial $f_{7}$ defined as

  $$\begin{aligned} f_{7} = x_1 x_2 x_3 x_4 x_5 x_6 x_7. \end{aligned}$$

• The space $\mathop {\mathrm {Harm}}\nolimits _8({\mathbb {R}}^d)^{D_d}$ is spanned by polynomials $f_{8,1}$, $f_{8,2}$, $f_{8, 3}$ respectively, as

  $$\begin{aligned} f_{8,1}&= \text {sym}(x_1^8)-\frac{28}{d-1}\text {sym}(x_1^2 x_2^6)+\frac{70}{d-1}\text {sym}(x_1^4 x_2^4), \quad d \ge 4, \\ f_{8,2}&= \text {sym}(x_1^4 x_2^4)-\frac{6}{d-2}\text {sym}(x_1^2 x_2^2 x_3^4) \\&\qquad +\frac{108}{(d-2)(d-3)} \text {sym}(x_1^2 x_2^2 x_3^2 x_4^2), \quad d \ge 4, \\ f_{8, 3}&= \text {sym}(x_1^5 x_2 x_3 x_4) - \frac{10}{9} \text {sym}(x_1^3 x_2^3 x_3 x_4), \quad d = 4, \\ f_{8, 4}&= x_1 x_2 x_3 x_4 x_5 x_6 x_7 x_8, \quad d = 8. \end{aligned}$$

• The space $\mathop {\mathrm {Harm}}\nolimits _9({\mathbb {R}}^d)^{D_d}$ is spanned by polynomials $f_{9,1}$ and $f_{9,2}$ defined as

  $$\begin{aligned} f_{9,1} =&\text {sym}(x_1^5 x_2 x_3 x_4 x_5) -\frac{5}{6} \text {sym}(x_1^3 x_2^3 x_3 x_4 x_5) , \quad d=5. \\ f_{9,2} =&x_1 x_2 x_3 x_4 x_5 x_6 x_7 x_8 x_9, \quad d=9. \end{aligned}$$

• The space $\mathop {\mathrm {Harm}}\nolimits _{10}({\mathbb {R}}^d)^{D_d}$ is spanned by polynomials $f_{10, 1}$, $f_{10, 2}$, $f_{10,3}$, $f_{10, 4}$ and $f_{10, 5}$ respectively, as

  $$\begin{aligned} f_{10,1}&= \text {sym}(x_{1}^{10}) - \frac{45}{d-1} \text {sym}(x_{1}^{2} x_{2}^{8}) + \frac{42}{d-1} \text {sym}(x_{1}^{4} x_{2}^{6}) \\&\quad + \frac{1008}{(d-1)(d-2)} \text {sym}(x_{1}^{2} x_{2}^{2} x_{3}^{6}) \\&\quad - \frac{1260}{(d-1)(d-2)} \text {sym}(x_{1}^{2} x_{2}^{4} x_{3}^{4}), \quad d \ge 4, \\ f_{10,2}&= \text {sym}(x_{1}^{4} x_{2}^{6}) - \frac{6}{d-2} \text {sym}(x_{1}^{2} x_{2}^{2} x_{3}^{6}) - \frac{30}{d-2} \text {sym}(x_{1}^{2} x_{2}^{4} x_{3}^{4}) \\&\quad + \frac{450}{(d-2)(d-3)} \text {sym}(x_{1}^{2}x_{2}^{2}x_{3}^{2}x_{4}^{4}) \\&\quad - \frac{10800}{(d-2)(d-3)(d-4)} \text {sym}(x_{1}^{2} x_{2}^{2} x_{3}^{2} x_{4}^{2} x_{5}^{2}) \quad d \ge 5, \\ f_{10,3}&= \text {sym}(x_1^7 x_2 x_3 x_4) -\frac{7}{3} \text {sym}(x_1^5 x_2^3 x_3 x_4) + \frac{70}{9} \text {sym}(x_1^3 x_2^3 x_3^3 x_4) , \quad d=4, \\ f_{10,4}&= \text {sym}(x_1^5 x_2 x_3 x_4 x_5 x_6) -\frac{2}{3} \text {sym}(x_1^3 x_2^3 x_3 x_4 x_5 x_6) , \quad d=6, \\ f_{10,5}&= x_1 x_2 x_3 x_4 x_5 x_6 x_7 x_8 x_9 x_{10} , \quad d=10. \end{aligned}$$

• The space $\mathop {\mathrm {Harm}}\nolimits _{11}({\mathbb {R}}^d)^{D_d}$ is spanned by polynomials $f_{11, 1}$, $f_{11, 2}$,and $f_{11, 3}$ respectively, as

  $$\begin{aligned} f_{11,1}&= \text {sym}(x_1^5 x_2^3 x_3 x_4 x_5) - \frac{4}{7} \text {sym}(x_1^7 x_2 x_3 x_4 x_5) - \frac{20}{9} \text {sym}(x_1^3 x_2^3 x_3^3 x_4 x_5) , \quad d=5, \\ f_{11,2}&= \text {sym}(x_1^5 x_2 x_3 x_4 x_5 x_6 x_7) - \frac{5}{9} \text {sym}(x_1^3 x_2^3 x_3 x_4 x_5 x_6 x_7) , \quad d=7, \\ f_{11,3}&= x_1 x_2 x_3 x_4 x_5 x_6 x_7 x_8 x_9 x_{10} x_{11}, \quad d=11. \end{aligned}$$

Substituting ${\varvec{v}}_k$ and ${\varvec{v}}_{k,a}$ for $D_d$-invariant harmonic polynomials

• For $d \ge 5$

  $$\begin{aligned} f_{4, 1} ({\varvec{v}}_k)&= {\left\{ \begin{array}{ll} \displaystyle \frac{1}{k} \left( 1 - 3\frac{k - 1}{d - 1}\right) , &{} 1 \le k \le d - 2, \\ \displaystyle - \frac{2}{d}, \; k = d - 1, d. \end{array}\right. } \\ f_{4,1} ({\varvec{v}}_{k,a})&= {\left\{ \begin{array}{ll} \displaystyle \frac{1}{(a^2 + k)^2} \left( a^4 + k - \frac{6}{d - 1} \left( k a^2 + \frac{k (k - 1)}{2} \right) \right) , \quad 1 \le k \le d - 3, \\ \displaystyle \frac{5 - 6 a^2 + a^4 - 2d}{(a^2 + d - 1)^2}, \; k = d - 2, d - 1. \end{array}\right. } \end{aligned}$$

  For $d = 4$

  $$\begin{aligned} f_{4, 2}({\varvec{v}}_1) = f_{4, 2}({\varvec{v}}_2) = 0, \quad f_{4, 2}({\varvec{v}}_3) = - \frac{1}{16}, \quad f_{4, 2}({\varvec{v}}_4) = \frac{1}{16}. \end{aligned}$$
  $$\begin{aligned} f_{4, 2} ({\varvec{v}}_{1, a}) = 0, \quad f_{4, 2} ({\varvec{v}}_{2, a}) = - \frac{a}{(a^2 + 3)^2}, \quad f_{4, 2} ({\varvec{v}}_{3, a}) = \frac{a}{(a^2 + 3)^2}. \end{aligned}$$

• For $d = 5$

  $$\begin{aligned} f_5({\varvec{v}}_1)= & {} f_5({\varvec{v}}_2) = f_5({\varvec{v}}_3) = 0, \quad f_5({\varvec{v}}_4) = - \frac{1}{25 \sqrt{5}}, \quad f_5({\varvec{v}}_5) = \frac{1}{25 \sqrt{5}}. \\ f_{5}({\varvec{v}}_{1, a})= & {} f_5 ({\varvec{v}}_{2, a}) = 0, \quad f_{5}({\varvec{v}}_{3, a}) = - \frac{a}{(a^2 + 4)^{5/2}}, \quad f_{5}({\varvec{v}}_{4, a}) = \frac{a}{(a^2 + 4)^{5/2}}. \end{aligned}$$

• For $d \ge 4$, $d \ne 6$

  $$\begin{aligned} f_{6, 1}({\varvec{v}}_k)&= {\left\{ \begin{array}{ll} \displaystyle \frac{1}{k^2} \left( 1 - 15 \frac{k - 1}{d - 1} + 30 \frac{(k - 1)(k -2)}{(d - 1)(d - 2)} \right) , \quad 1 \le k \le d - 2, \\ \displaystyle \frac{16}{d^2}, \quad k = d - 1, d. \end{array}\right. } \\ f_{6,1} ({\varvec{v}}_{k,a})&= {\left\{ \begin{array}{ll} \displaystyle \frac{1}{(a^2 + k)^3} \bigg (a^6 + k - \frac{15}{d - 1} \left( a^2 k + a^4 k + k (k - 1) \right) \\ \displaystyle \qquad + \frac{180}{(d - 1) (d - 2)} \left( a^2 \frac{k (k - 1)}{2} + \frac{k (k - 1) (k - 2)}{3!} \right) \bigg ), \quad 1 \le k \le d- 3, \\ \displaystyle \frac{-61 + 75 a^2 - 15 a^4 + a^6 + 16 d}{(a^2 + d - 1)^3}, \quad k = d - 2, d - 1. \end{array}\right. } \end{aligned}$$

  For $d = 6$

  $$\begin{aligned} f_{6, 2}({\varvec{v}}_1)= & {} \dots = f_{6, 2}({\varvec{v}}_4) = 0, \quad f_{6, 2}({\varvec{v}}_5) = - \frac{1}{216}, \quad f_{6, 2}({\varvec{v}}_6) = \frac{1}{216}. \\ f_{6,2} ({\varvec{v}}_{a,1})= & {} f_{6,2} ({\varvec{v}}_{a,2}) = f_{6,2} ({\varvec{v}}_{a,3}) = 0, \quad \\ f_{6,2} ({\varvec{v}}_{a,4})= & {} - \frac{a}{(a^2 + 5)^3}, \quad f_{6,2} ({\varvec{v}}_{a,4}) = \frac{a}{(a^2 + 5)^3}. \end{aligned}$$

• For $d = 7$

  $$\begin{aligned} f_7({\varvec{v}}_1)= & {} \cdots = f_7 ({\varvec{v}}_5) = 0, \quad f_7({\varvec{v}}_6) = - \frac{1}{343\sqrt{7}}, \quad f({\varvec{v}}_7) = \frac{1}{343\sqrt{7}}. \\ f_{7}({\varvec{v}}_{1,a})= & {} \cdots = f_7({\varvec{v}}_{5,a}) = 0, \quad f_{7}({\varvec{v}}_{6,a}) = - \frac{a}{(a^2 + 6)^{7/2}},\\ f_{7}({\varvec{v}}_{6,a}) =& - \frac{a}{(a^2 + 6)^{7/2}}. \end{aligned}$$

• For $d \ge 2$

  $$\begin{aligned} f_{8,1} ({\varvec{v}}_k)&= {\left\{ \begin{array}{ll} \displaystyle \frac{1}{k^3} \left( 1 + 7 \frac{k - 1}{d - 1} \right) , \quad 1 \le k \le d - 2, \\ \displaystyle \frac{8}{d^3}, \quad k = d - 1, d. \end{array}\right. } \\ f_{8,1} ({\varvec{v}}_{k,a})&= {\left\{ \begin{array}{ll} \displaystyle \frac{1}{(a^2 + k)^4} \bigg (a^8 + k - \frac{28}{d - 1} (a^2 k + a^6 k + k (k - 1)) \\ \displaystyle \qquad + \frac{70}{d - 1}\left( a^4 k + \frac{k (k - 1)}{2} \right) \bigg ), \quad 1 \le k \le d - 3, \\ \displaystyle \frac{-15 - 28 a^2 + 70 a^4 - 28 a^6 + a^8 + 8 d}{(a^2 + d - 1)^4}, \quad k = d - 2, d - 1. \end{array}\right. } \end{aligned}$$

  For $d \ge 4$

  $$\begin{aligned} f_{8,2} ({\varvec{v}}_k)&= {\left\{ \begin{array}{ll} \displaystyle \frac{k - 1}{2 k^3} \left( 1 - 6 \frac{k - 2}{d - 2} + 9 \frac{(k - 2) (k - 3)}{(d - 2)(d - 3)} \right) , \quad 1 \le k \le d - 2, \\ \displaystyle \frac{2(d - 1)}{d^3}, \quad k = d - 1, d. \end{array}\right. } \\ f_{8,2} ({\varvec{v}}_{a,k})&= {\left\{ \begin{array}{ll} \displaystyle \frac{1}{(a^2 + k)^4} \bigg (a^4 k + \frac{(k-1) k}{2} - \frac{6}{d -2} \displaystyle \bigg (a^2 (k -1) k + \frac{a^2 (k-1) k}{2} \\ \qquad + \frac{(k - 2) (k - 1) k}{2} \bigg ) + \frac{9 k (-3 + 4 a^2 + k) (k - 2)(k - 1)}{2 (d - 3) (d - 2)} \bigg ), \quad 1 \le k \le d - 3, \\ \displaystyle \frac{- 2 (5 - 6 a^2 + a^4 - d) (-1 + d)}{(a^2 + d - 1)^4}, \quad k = d - 2, d - 1. \end{array}\right. } \end{aligned}$$

  For $d = 4$

  $$\begin{aligned} f_{8,3} ({\varvec{v}}_1)= & {} f_{8, 3} ({\varvec{v}}_2) = 0, \quad f_{8,3} ({\varvec{v}}_3) = - \frac{1}{96}, \quad f_{8,3} ({\varvec{v}}_4) = \frac{1}{96}. \\ f_{8, 3} ({\varvec{v}}_{a,a})= & {} 0, \quad f_{8, 3} ({\varvec{v}}_{2,a}) = \frac{a + 10 a^3 - 3 a^5}{3 (3 + a^2)^4}, \quad f_{8, 3} ({\varvec{v}}_{3,a}) = \frac{-a - 10 a^3 + 3 a^5}{3 (3 + a^2)^4}. \end{aligned}$$

  For $d = 8$,

  $$\begin{aligned} f_{8, 4}({\varvec{v}}_1)= & {} \cdots f_{8,4}({\varvec{v}}_6) = 0, \quad f_{8, 4}({\varvec{v}}_7) = - \frac{1}{4096}, \quad f_{8, 4}({\varvec{v}}_8) = \frac{1}{4096}. \\ f_{8, 4}({\varvec{v}}_{1,a})= & {} \cdots f_{8,4}({\varvec{v}}_{6,a}) = 0, \quad f_{8, 4}({\varvec{v}}_{7,a}) = - \frac{a}{(a^2 + 7)^4}, \quad f_{8, 4}({\varvec{v}}_{8,a}) = \frac{a}{(a^2 + 7)^4}. \end{aligned}$$

• For $d=5$

  $$\begin{aligned} f_{9,1} ({\varvec{v}}_{1})= & {} f_{9, 1} ({\varvec{v}}_{2}) = f_{9,1} ({\varvec{v}}_{3}) = 0, \quad f_{9, 1} ({\varvec{v}}_{4}) = \frac{2}{375 \sqrt{5}}, \quad f_{9,1} ({\varvec{v}}_{5}) = - \frac{2}{375 \sqrt{5}}. \\ f_{9,1}({\varvec{v}}_{1,a})= & {} f_{9,1}({\varvec{v}}_{2,a}) = 0, \quad f_{9,1}({\varvec{v}}_{3,a}) = \frac{1}{(a^2 + 4)^{\frac{9}{2}}} \left( -a^5 + \frac{10}{3} a^3 + a \right) , \quad \\ f_{9,1}({\varvec{v}}_{4,a})= & {} \frac{1}{(a^2 + 4)^{\frac{9}{2}}} \left( a^5 - \frac{10}{3} a^3 - a \right) . \end{aligned}$$

  For $d = 9$

  $$\begin{aligned} f_{9,2} ({\varvec{v}}_1)= & {} \cdots = f_{9, 2} ({\varvec{v}}_7) = 0, \quad f_{9,2} ({\varvec{v}}_{8}) = - \frac{1}{19683}, \quad f_{9,2} ({\varvec{v}}_{9}) = \frac{1}{19683}. \\ f_{9,2}({\varvec{v}}_{1,a})= & {} \cdots = f_{9,2}({\varvec{v}}_{6,a}) = 0, \quad f_{9,2}({\varvec{v}}_{7,a}) = - \frac{a}{(a^2 + 8)^{\frac{9}{2}}}, \quad \\ f_{9,2}({\varvec{v}}_{8,a})= & {} \frac{a}{(a^2 + 8)^{\frac{9}{2}}}. \end{aligned}$$

• For $d \ge 4$

  $$\begin{aligned} f_{10,1}({\varvec{v}}_{k})&= {\left\{ \begin{array}{ll}\!\frac{1}{k^5} \left( k - \frac{3k(k-1)}{d-1} - \frac{126k(k-1)(k-2)}{(d-1)(d-2)} \right) , &{} 1 \le k \le d-2, \\\! \frac{1}{d^5} \left( k-1 - \frac{3k(k+1)}{d-1} - \frac{126k(k+1)(k-1)}{(d-1)(d-2)} \right) , & k=d-1, \\\! \frac{1}{d^5} \left( k - \frac{3k(k-1)}{d-1} - \frac{126k(k-1)(k-2)}{(d-1)(d-2)} \right) , \; k=d. \end{array}\right. } \\ g_{10,1}({\varvec{v}}_{k,a})&= {\left\{ \begin{array}{ll} \frac{1}{(a^2 + k)^5} \Biggl ( a^{10} + k - \frac{45}{d-1} \left( a^2 k + a^8 k + k(k-1)\right) + \frac{42}{d-1} \left( a^4 k + a^6 k + k(k-1) \right) \\ \quad + \frac{1008}{(d-1)(d-2)} \left( \frac{a^6 k(k-1)}{2} + a^2 k(k-1) + \frac{k(k-1)(k-2)}{2} \right) \\ \quad - \frac{1260}{(d-1)(d-2)} \left( \frac{a^2 k (k-1)}{2} + a^4 k(k-1) + \frac{k(k-1)(k-2)}{2} \right) \Biggr ), \; 1 \le k \le d-3, \\ \frac{1}{(a^2 + d-1)^5} \Biggl ( a^{10} + k - \frac{45}{d-1} \left( a^8 k + a^2 k + k^2 - k \right) + \frac{42}{d-1} \left( a^6 k + a^4 k + k^2 -k \right) \\ \quad + \frac{1008}{(d-1)(d-2)} \Bigl ( a^6 (k-1) + \frac{a^6 (k-1)(k-2)}{2} + \frac{(k-1)(k-2)}{2} + 2a^2 (k-1) + a^2 (k-1)(k-2) \\ \quad + (k-1)(k-2) + \frac{(k-1)(k-2)(k-3)}{2} \Bigr ) - \frac{1260}{(d-1)(d-2)} \Bigl ( a^2(k-1) + \frac{a^2 (k-1)(k-2)}{2} + \frac{(k-1)(k-2)}{2} \\ \quad + 2a^4 (k-1) + a^4 (k-1)(k-2) + (k-1)(k-2) + \frac{(k-1)(k-2)(k-3)}{2} \Bigr ) \Biggr ), \; k=d-2, \\ \frac{1}{(a^2 + d-1)^5} \Biggl ( a^{10} + k - 1 - \frac{45}{d-1} \Bigl ( a^2(k-1) + a^8(k-1) + (k-1)(k-2) \Bigr ) \\ \quad + \frac{42}{d-1} \Bigl ( a^4(k-1) + a^6(k-1) + (k-1)(k-2) \Bigr ) + \frac{1008}{(d-1)(d-2)} \Bigl ( \frac{a^6(k-1)(k-2)}{2} + a^2(k-1)(k-2) \\ \quad + \frac{(k-1)(k-2)(k-3)}{2} \Bigr ) - \frac{1260}{(d-1)(d-2)} \Bigl ( \frac{a^2(k-1)(k-2)}{2} + a^4(k-1)(k-2) + \frac{(k-1)(k-2)(k-3)}{2} \Bigr ) \Biggr ), \; k=d -1. \end{array}\right. } \end{aligned}$$

• For $d \ge 5$

  $$\begin{aligned} f_{10,2}({\varvec{v}}_{k})&= {\left\{ \begin{array}{ll} \frac{1}{k^5} \Bigl ( k(k-1) - \frac{18k(k-1)(k-2)}{n-2} + \frac{75k(k-1)(k-2)(k-3)}{(n-2)(n-3)} - \frac{90k(k-1)(k-2)(k-3)(k-4)}{(n-2)(n-3)(n-4)} \Bigr ), \; 1 \le k \le d-2, \\ \frac{1}{n^5} \Biggl ( k(k+1) - \frac{18k(k+1)(k-1)}{n-2} + \frac{450}{(n-2)(n-3)}\Bigl ( \frac{k(k-1)(k-2)}{2} + \frac{k(k-1)(k-2)(k-3)}{3!} + \frac{k(k-1)(k-2)}{3!} \Bigr ) \\ \quad - \frac{10800}{(n-2)(n-3)(n-4)} \left( \frac{k(k-1)(k-2)(k-3)}{4!} + \frac{k(k-1)(k-2)(k-3)(k-4)}{5!} \right) \Biggr ), \; k=d-1, \\ \frac{1}{n^5} \Bigl ( k(k-1) - \frac{18k(k-1)(k-2)}{n-2} + \frac{75k(k-1)(k-2)(k-3)}{(n-2)(n-3)} - \frac{90k(k-1)(k-2)(k-3)(k-4)}{(n-2)(n-3)(n-4)} \Bigr ), \; k=d. \end{array}\right. } \\ g_{10,2}({\varvec{v}}_{k,a})&= {\left\{ \begin{array}{ll} \frac{1}{(a^2 + k)^5} \Biggl ( a^4 k + a^6 k + k(k-1) - \frac{6}{d-2} \Bigl ( \frac{a^6 k(k-1)}{2} + a^2 k(k-1)+ \frac{k(k-1)(k-2)}{2} \Bigr ) \\ \quad - \frac{30}{d-2} \left( \frac{a^2 k(k-1)}{2} + a^4 k(k-1) + \frac{k(k-1)(k-2)}{2} \right) + \frac{450}{(d-2)(d-3)} \Bigl ( \frac{a^4 k(k-1)(k-2)}{3!} \\ \quad + \frac{a^2 k(k-1)(k-2)}{2}+ \frac{k(k-1)(k-2)(k-3)}{3!} \Bigr ) \\ \quad - \frac{10800}{(d-2)(d-3)(d-4)} \Bigl ( \frac{a^2 k(k-1)(k-2)(k-3)}{4!} + \frac{k(k-1)(k-2)(k-3)(k-4)}{5!} \Bigr ) \Biggr ), \; 1 \le k \le d - 3, \\ \frac{1}{(a^2 + d-1)^5} \Biggl ( a^4 k + a^6 k + k^2 -k - \frac{6}{d-2} \Bigl ( a^6(k-1) + \frac{a^6(k-1)(k-2)}{2} + 2a^2(k-1) \\ \quad + \frac{(k-1)(k-2)}{2} + a^2(k-1)(k-2) + (k-1)(k-2) + \frac{(k-1)(k-2)(k-3)}{2} \Bigr ) \\ \quad - \frac{30}{d-2} \Bigl ( a^2(k-1) + \frac{a^2(k-1)(k-2)}{2} + 2a^4(k-1) + \frac{(k-1)(k-2)}{2} + a^4(k-1)(k-2) \\ \quad + (k-1)(k-2) + \frac{(k-1)(k-2)(k-3)}{2} \Bigr ) + \frac{450}{(d-2)(d-3)} \Bigl ( \frac{a^4(k-1)(k-2)}{2} \\ \quad + \frac{a^4(k-1)(k-2)(k-3)}{3!} + \frac{a^2(k-1)(k-2)}{2} + \frac{(k-1)(k-2)(k-3)}{3!} + a^2(k-1)(k-2) \\ \quad + \frac{a^2(k-1)(k-2)(k-3)}{2} + \frac{(k-1)(k-2)(k-3)}{2} + \frac{(k-1)(k-2)(k-3)(k-4)}{3!} \Bigr ) \\ \quad - \frac{10800}{(d-2)(d-3)(d-4)} \Bigl ( \frac{a^2(k-1)(k-2)(k-3)}{3!} + \frac{a^2(k-1)(k-2)(k-3)(k-4)}{4!} \\ \quad + \frac{(k-1)(k-2)(k-3)(k-4)}{4!} + \frac{(k-1)(k-2)(k-3)(k-4)(k-5)}{5!} \Bigr ) \Biggr ), \; k=d-2, \\ \frac{1}{(a^2 + d - 1)^5} \Biggl ( a^4(k-1) + a^6(k-1) + (k-1)(k-2) - \frac{6}{d-2} \Bigl ( \frac{a^6(k-1)(k-2)}{2} + a^2(k-1)(k-2) \\ \quad + \frac{(k-1)(k-2)(k-3)}{2} \Bigr ) - \frac{30}{d-2} \Bigl ( \frac{a^2(k-1)(k-2)}{2} + a^4(k-1)(k-2) + \frac{(k-1)(k-2)(k-3)}{2} \Bigr ) \\ \quad + \frac{450}{(d-2)(d-3)} \Bigl ( \frac{a^4(k-1)(k-2)(k-3)}{3!} + \frac{a^2(k-1)(k-2)(k-3)}{2} + \frac{(k-1)(k-2)(k-3)(k-4)}{3!} \Bigr ) \\ \quad - \frac{10800}{(d-2)(d-3)(d-4)} \Bigl ( \frac{a^2(k-1)(k-2)(k-3)(k-4)}{4!} + \frac{(k-1)(k-2)(k-3)(k-4)(k-5)}{5!} \Bigr ) \Biggr ),\; k=d-1. \end{array}\right. } \end{aligned}$$

• For $d = 4$

  $$\begin{aligned} f_{10,3} ({\varvec{v}}_{1})&= f_{10, 3} ({\varvec{v}}_{2})= 0, \quad f_{10,3} ({\varvec{v}}_{3}) = - \frac{1}{144}, \quad f_{10,3} ({\varvec{v}}_{4}) = \frac{1}{144}. \\ f_{10,3} ({\varvec{v}}_{1,a})&= 0, \quad f_{10,3} ({\varvec{v}}_{2,a}) = \frac{1}{(a^2+3)^5} \left( -a^7 + 7a^5 - \frac{49}{3}a^3 + \frac{29}{9}a \right) , \\ f_{10,3} ({\varvec{v}}_{3,a})&= \frac{1}{(a^2+3)^5} \left( a^7 - 7a^5 + \frac{49}{3}a^3 - \frac{29}{9}a \right) . \end{aligned}$$

• For $d=6$

  $$\begin{aligned} f_{10,4}({\varvec{v}}_{1})&= \cdots = f_{10, 4} ({\varvec{v}}_{4}) = 0, \quad f_{10, 4}({\varvec{v}}_{5}) = \frac{1}{1944}, \quad f_{10, 4} ({\varvec{v}}_6) = - \frac{1}{1944}. \\ f_{10,4} ({\varvec{v}}_{1,a})&= \cdots = f_{10,4} ({\varvec{v}}_{3,a}) = 0, \quad f_{10,4} ({\varvec{v}}_{4,a}) = \frac{1}{(a^2+5)^5} \left( - a^5 + \frac{10}{3}a^3 + \frac{5}{3}a \right) , \quad \\ f_{10,4} ({\varvec{v}}_{5, a})&= \frac{1}{(a^2+5)^5} \left( a^5 - \frac{10}{3}a^3 - \frac{5}{3}a \right) . \end{aligned}$$

• For $d=10$

  $$\begin{aligned} f_{10,5} ({\varvec{v}}_{1})&= \cdots f_{10,5} ({\varvec{v}}_{8})= 0, \quad f_{10,5} ({\varvec{v}}_{9}) = - \frac{1}{100000}, \quad f_{10,5} ({\varvec{h}}_{k}) = \frac{1}{100000}. \\ f_{10,5} ({\varvec{v}}_{1, a})&= \cdots f_{10,5} ({\varvec{v}}_{7,a}) = 0, \quad f_{10,5} ({\varvec{v}}_{8,a}) = - \frac{a}{(a^2+9)^5}, \quad \\ f_{10,5} ({\varvec{v}}_{9,a})&= \frac{a}{(a^2+9)^5}. \end{aligned}$$

• For $d=5$

  $$\begin{aligned} f_{11,1} ({\varvec{v}}_{1})&= \cdots = f_{11,1}({\varvec{v}}_3) = 0, \quad f_{11,1} ({\varvec{v}}_{4}) = \frac{64}{39375 \sqrt{5}}, \quad f_{11,1} ({\varvec{v}}_{5}) = - \frac{64}{39375 \sqrt{5}}. \\ f_{11,1} ({\varvec{v}}_{1,a})&= f_{11,1} ({\varvec{v}}_{2,a})= 0, \quad f_{11,1} ({\varvec{v}}_{3,a}) = \frac{1}{(a^2+4)^{\frac{11}{2}}} \left( \frac{4}{7}a^7 - 4a^5 + \frac{28}{3}a^3 - \frac{52}{63}a \right) , \\ f_{11,1} ({\varvec{v}}_{4,a})&= \frac{1}{(a^2+4)^{\frac{11}{2}}} \left( - \frac{4}{7}a^7 + 4a^5 - \frac{28}{3}a^3 + \frac{52}{63}a \right) . \end{aligned}$$

• For $d = 7$

  $$\begin{aligned} f_{11,2} ({\varvec{v}}_{1})&= \cdots = f_{11,2} ({\varvec{v}}_{5})= 0, \quad f_{11,2} ({\varvec{v}}_{6}) = \frac{2}{7203 \sqrt{7}}, \quad f_{11,2} ({\varvec{v}}_{7}) = - \frac{2}{7203 \sqrt{7}}. \\ f_{11,2} ({\varvec{v}}_{1,a})&= \cdots = f_{11,4}({\varvec{v}}_{4,a}) = 0, \quad f_{11,2} ({\varvec{v}}_{5,a}) = \frac{1}{(a^2 + 6)^{\frac{11}{2}}} \left( -a^5 + \frac{10}{3}a^3 + \frac{7}{3}a \right) , \\ f_{11,2} ({\varvec{v}}_{6,a})&= \frac{1}{(a^2 + 6)^{\frac{11}{2}}} \left( a^5 - \frac{10}{3}a^3 - \frac{7}{3}a \right) . \end{aligned}$$

• For $d=11$

  $$\begin{aligned} f_{11,3} ({\varvec{v}}_{k})&= \cdots = f_{11,3}({\varvec{v}}_{9})= 0, \quad f_{11,3} ({\varvec{v}}_{10}) = - \frac{1}{161051 \sqrt{11}}, \quad \\ f_{11,3} ({\varvec{v}}_{11})&= \frac{1}{161051 \sqrt{11}}. \\ f_{11,3} ({\varvec{v}}_{1,a})&= \cdots = f_{11,3} ({\varvec{v}}_{8,a}) = 0, \quad f_{11,3} ({\varvec{v}}_{9,a}) = - \frac{a}{(a^2 + 10)^{\frac{11}{2}}}, \quad \\ f_{11,3} ({\varvec{v}}_{10,a})&= \frac{a}{(a^2 + 10)^{\frac{11}{2}}}. \end{aligned}$$