# Functions on the Plane as Combinations of Powers of Distances to Points

Authors: Jose L. Martinez-Morales
Journal: Abstract and Applied Analysis
 (2010)

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## Abstract

Some real functions on the plane can be expressed as a linear combination of powers of distances to certain points.


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## Body

## 1. Introduction

In this paper we construct a function basis to express different powers of the distance between a point and the origin. Namely, for two given natural numbersm and n we find points in the plane such that the function rn+2mcos(nθ) can be expressed as linear combination of powers of distances to those points (see  formula (3.15) below), both the linear combination and the given function having the same domain in the plane (for the notion of special function that we use, see Section 3.1).One of the important observations in the theory of real functions is that series expansions lead to certain interesting numbers and functions (e.g., Fourier coefficients,L-functions). Namely, usually series representations can be constructed explicitly in some model spaces of sections of various function bundles over appropriate analytic manifolds. (By considering functions on the plane as maps F: ℝ2→ℝ, the graphs of these functions are the sections described by the map ψ: ℝ2→ℝ3; (x,y) ↦ (x,y, F(x,y)), which defines 2-sub-manifolds on ℝ3.) (for analytic functions on the plane ℝ2; see Section 2.2).We denote by𝒜(ℝ2) the subspace of real analytic functions. It is well known that any polynomial of degree n can be written exactly as a linear combination of powers of the right number of terms of the form x·vk+bk. We realize a series representation in the space of analytic functions on the plane 𝒜(ℝ2).Consider a pointp in the plane ℝ2 with coordinates (rcosθ,rsinθ).The distance used here is the Euclidean distance. Denote(1.1)|p|=r.We prove the following theorem.Theorem 1.1.
Consider two natural numbersm and n. There are N numbers a1,…,aN and N points p1,…,pN, such that
(1.2)rn+2mcos⁡(nθ)=∑i=1Nai|p-pi|2(m+n).Remark 1.2.
The numbersa1,…,aN and the points  p1,…,pN can be determined exactly.Corollary 1.3.
The linear combinations ofrn+2mcos(nθ), m, n=0,1,…, can be expressed similarly.Remark 1.4.
An analogous theorem holds for the sine function.Denote byx̲ the maximum integer less than or equal to x.Consider the function(1.3)δ(x,y)={0ifx≠y,1ifx=y.


## 2. Proof of Theorem1.1

We will denote bypd the Legendre polynomial of degree d.The following formula is formula (8.921) of [1]:(2.1)11-2tx+t2=∑d=0∞tdpd(x),|t|<min⁡|x±x2-1|.
Consider the set |t|<min|x±x2-1| of ℝ2. Consider a subset ℋ (not necessarily proper) of such set. We will work solely with analytic functions on ℋ.Consider the series∑i=0∞∑j=0iaijti+jxi-j.Divide this number by1-2tx+t2. By formula (2.1), the following formula holds:(2.2)∑i=0∞∑j=0iaijti+jxi-j1-2tx+t2=∑i=0∞∑j=0iaijti+jxi-j⋅∑d=0∞tdpd(x).
We use the formula of the product of two series ∑n=0∞an and ∑n=0∞bn:(2.3)∑n=0∞an∑n=0∞bn=∑n=0∞∑k=0nakbn-k.
The right side of formula (2.2) becomes(2.4)∑i=0∞∑j=0iaijti+jxi-j⋅∑d=0∞tdpd(x)=∑i=0∞∑k=0i∑j=0i-kai-k,jpk(x)ti+jxi-k-j.
Let us change the order of addition with respect to the indices j and k:(2.5)∑k=0i∑j=0i-kai-k,jpk(x)ti+jxi-k-j=∑j=0i∑k=0i-jai-k,jpk(x)ti+jxi-k-j.
The following formula is formula (8.911) of [1]:(2.6)pk(x)=∑l=0k/2̲bklxk-2l,
where(2.7)bkl=12k(-1)l(2k-2l)!l!(k-l)!(k-2l)!.
Substituting (2.6) in (2.5),(2.8)∑j=0i∑k=0i-jai-k,jpk(x)ti+jxi-k-j=∑j=0i∑k=0i-jai-k,j∑l=0k/2̲bklxk-2lti+jxi-k-j=∑j=0i∑k=0i-j∑l=0k/2̲ai-k,jbklti+jxi-j-2l.
Let us change the order of addition with respect to the indices k and l:(2.9)∑k=0i-j∑l=0k/2̲ai-k,jbklti+jxi-j-2l=∑l=0(i-j)/2̲∑k=2li-jai-k,jbklti+jxi-j-2l.
Denote by(2.10)aijl=∑k=2li-jai-k,jbkl.
The sum (2.9) is ∑l=0(i-j)/2̲aijlti+jxi-j-2l. We will substitute this sum in (2.8).(2.11)∑j=0i∑k=0i-j∑l=0k/2̲ai-k,jbklti+jxi-j-2l=∑j=0i∑l=0(i-j)/2̲aijlti+jxi-j-2l.Let us change the order of addition with respect to the indicesj and l:(2.12)∑j=0i∑l=0(i-j)/2̲aijlti+jxi-j-2l=∑l=0i/2̲∑j=0i-2laijlti+jxi-j-2l.
Let us add with respect to the index i from zero to infinity(2.13)∑i=0∞∑l=0i/2̲∑j=0i-2laijlti+jxi-j-2l.
Let us change the order of addition with respect to the indices i and l:(2.14)∑i=0∞∑l=0i/2̲∑j=0i-2laijlti+jxi-j-2l=∑l=0∞∑i=2l∞∑j=0i-2laijlti+jxi-j-2l.
Change the index i for index i′=i-2l:(2.15)∑l=0∞∑i=2l∞∑j=0i-2laijlti+jxi-j-2l=∑l=0∞∑i′=0∞∑j′=0i′ai′+2l,j′lti′+2l+j′xi′-j′.
We change the indices i′ and j′ for the indices (2.16)i′=i-l,j′=j-l,∑l=0∞∑i′=0∞∑j′=0i′ai′+2l,j′lti′+2l+j′xi′-j′=∑l=0∞∑i=l∞∑j=li-lai+l,j-l,lti+jxi-j.
Let us change the order of addition with respect to the indices i and l:(2.17)∑l=0∞∑i=l∞∑j=li-lai+l,j-l,lti+jxi-j=∑i=0∞∑l=0i∑j=li-lai+l,j-l,lti+jxi-j.
Let us change the order of addition with respect to the indices j and l:(2.18)∑l=0i∑j=li-lai+l,j-l,lti+jxi-j=∑j=0i∑l=0jai+l,j-l,lti+jxi-j.
Denote by(2.19)Aij=∑l=0jai+l,j-l,l.
The sum (2.18) becomes(2.20)∑l=0i∑j=li-lai+l,j-l,lti+jxi-j=∑j=0iAijti+jxi-j.
Let us add with respect to the index i from zero to infinity.The ratio (2.2) becomes(2.21)∑i=0∞∑j=0iaijti+jxi-j1-2tx+t2=∑i=0∞∑j=0iAijti+jxi-j.


### 2.1. Coefficients with Three Indices

Let us calculate the numbersai+l,j-l,l. By formula (2.10), the following formula holds:(2.22)ai+l,j-l,l=∑k=2li+2l-jai+l-k,j-lbkl.
Change the index k for index k-2l:(2.23)∑k=2li+2l-jai+l-k,j-lbkl=∑k=0i-jai-l-k,j-lb2l+k,l.
By formula (2.7), the following formula holds:(2.24)b2l+k,l=(-1)l(2l+2k)!22l+kl!(l+k)!k!.
Let us use the formula(2.25)dndxn1x=(-1)n(2n)!4-nx-n-1/2n!
to write(2.26)(-1)l(2l+2k)!22l+kl!(l+k)!k!=(-1)k2kxl+k+1/2l!k!dl+kdxl+k1x.
We will substitute this in (2.23):(2.27)∑k=0i-jai-l-k,j-lb2l+k,l=∑k=0i-jai-l-k,j-l(-1)k2kxl+k+1/2l!k!dl+kdxl+k1x.
Formula (2.22) becomes(2.28)ai+l,j-l,l=∑k=0i-jai-l-k,j-l(-1)k2kxl+k+1/2l!k!dl+kdxl+k1x.


### 2.2. A Series of Half-Integer Powers

Consider three numbersa, b, and c.We will develop the power of the sum of three numbers.Lemma 2.1.
Consider a natural numbern. Then, the following formula holds:
(2.29)(a+b+c)n=∑i=0n∑j=0in!an-ibi-jcj(n-i)!j!(i-j)!.Consider the series∑n=0∞an(a2-2abc+c2)n.By Lemma2.1 the following formula holds:(2.30)∑n=0∞an(a2-2abc+c2)n=∑n=0∞an∑i=0n∑j=0i(-1)i-j2i-jn!a2n-i-jbi-jci+j(n-i)!j!(i-j)!.
We will distribute this sum, and we will exchange the order of addition:(2.31)∑n=0∞an(a2-2abc+c2)n=∑i=0∞∑j=0i∑n=i∞an(-1)i-j2i-jn!a2n-i-jbi-jci+j(n-i)!j!(i-j)!.
Denote by(2.32)aij=∑n=i∞an(-1)i-j2i-jn!a2n-i-j(n-i)!j!(i-j)!.
The series becomes(2.33)∑n=0∞an(a2-2abc+c2)n=∑i=0∞∑j=0iaijbi-jci+j.
Consider the formula(2.34)dixndxi|x=a2=a2n-2in!(n-i)!.


#### 2.2.1. Formulas for Coefficients with Two Indices

Let us use formula (2.34) to rewrite formula (2.32).(2.35)aij=∑n=i∞an(-1)i-j2i-jai-jj!(i-j)!dixndxi|x=a2.
Consider the function(2.36)f(x)=∑n=0∞anxn.
Let us differentiate the series term by term:(2.37)didxif(x)=∑n=0∞andixndxi.
The i-1 first terms in this series are equal to zero. Then, the following formula holds:(2.38)didxif(x)=∑n=i∞andixndxi.
The number aij becomes(2.39)aij=(-1)i-j2i-jai-jj!(i-j)!dif(x)dxi|x=a2.(1) Coefficients with Two Indices
We write the numbersai-l-k,j-l substituting the numbers aij for numbers ai+j-1aij.
By formula (2.39), the following formula holds:
(2.40)ai+j-k-2l-1ai-l-k,j-l=(-1)i-k-j2i-k-ja2i-2k-2l-1(j-l)!(i-k-j)!di-k-lf(x)dxi-k-l|x=a2.
Let us substitute in (2.28):
(2.41)ai+l,j-l,l=∑k=0i-j(-1)i-k-j2i-k-ja2i-2k-2l-1(j-l)!(i-k-j)!di-k-lf(x)dxi-k-l|x=a2(-1)k2kxl+k+1/2l!k!dl+kdxl+k1x.
Let
(2.42)x=a2.
Then, the following formula holds:
(2.43)ai+l,j-l,l=∑k=0i-j(-1)i-j2i-ja2i(j-l)!(i-k-j)!l!k!di-k-lf(x)dxi-k-l|x=a2dl+kdxl+k1x|x=a2.
The sum equals
(2.44)ai+l,j-l,l=(-1)i-j2i-ja2i(i-j)!(j-l)!l!di-jdxi-j(dj-lf(x)dxj-ldldxl1x)|x=a2.
Let us add with respect of index l from zero to j:
(2.45)∑l=0jai+l,j-l,l=∑l=0j(-1)i-j2i-ja2i(i-j)!(j-l)!l!di-jdxi-j(dj-lf(x)dxj-ldldxl1x)|x=a2=(-1)i-j2i-ja2i(i-j)!di-jdxi-j∑l=0j1(j-l)!l!dj-lf(x)dxj-ldldxl1x|x=a2.
The sum equals
(2.46)∑l=0j1(j-l)!l!dj-lf(x)dxj-ldldxl1x=1j!djdxjf(x)x.
We will substitute this sum in (2.45):
(2.47)∑l=0jai+l,j-l,l=(-1)i-j2i-ja2i(i-j)!di-jdxi-j1j!djdxjf(x)x|x=a2=(-1)i-j2i-ja2i(i-j)!j!didxif(x)x|x=a2.
We will substitute this sum in (2.19):
(2.48)Aij=(-1)i-j2i-ja2ij!(i-j)!didxif(x)x|x=a2.


#### 2.2.2. A Series of Half-Integer Powers

Divide both sides of (2.33) by a2-2abc+c2:(2.49)∑n=0∞an(a2-2abc+c2)na2-2abc+c2=∑i=0∞∑j=0iaijbi-jci+ja2-2abc+c2=∑i=0∞∑j=0iaijbi-jci+ja1-2(bc/a)+c2/a2.
By formula (2.21), the following formula holds:(2.50)∑i=0∞∑j=0iaijbi-jci+ja1-2(bc/a)+c2/a2=∑i=0∞∑j=0iAijci+jbi-j,
where  (i)
coefficientsaij in (2.21) are replaced by coefficients ai+j-1aij,(ii)
coefficientsAij are given by formula (2.48).The left side of (2.49) becomes(2.51)∑n=0∞an(a2-2abc+c2)na2-2abc+c2=∑i=0∞∑j=0iAijci+jbi-j.


### 2.3. A Series of Half-Integer Powers

Consider the series∑n=0∞anxn/2.Let us separate this sum into(i)
a sum with respect to even indices,(ii)
a sum over odd indices.Then, one has(2.52)∑n=0∞anxn/2=∑n=0∞a2nxn+∑n=0∞a1+2nx1/2+n=∑n=0∞a2nxn+∑n=0∞a1+2nxn+1x.
In the second sum, we change the index n for index n′=n+1:(2.53)∑n=0∞a1+2nxn+1=∑n′=1∞a-1+2n′xn′.
Define(2.54)a-1=0.
Then, the following formula holds:(2.55)∑n′=1∞a-1+2n′xn′=∑n=0∞a-1+2nxn.
Formula (2.52) becomes(2.56)∑n=0∞anxn/2=∑n=0∞a2nxn+∑n=0∞a-1+2nxnx.
We will denote  (2.57)f(x)=∑n=0∞a2nxn,g(x)=∑n=0∞a-1+2nxnx.
Let(2.58)x=a2-2abc+c2.
By formulas (2.33) and (2.51), the following formula holds:(2.59)∑n=0∞an(a2-2abc+c2)n/2=∑i=0∞∑j=0iaijbi-jci+j+∑i=0∞∑j=0ibijbi-jci+j,
where (see (2.39) and (2.48)) (2.60)aij=(-1)i-j2i-jai-jj!(i-j)!dif(x)dxi|x=a2,bij=(-1)i-j2i-ja2ij!(i-j)!dig(x)dxi|x=a2.
We will denote(2.61)Aij=(-1)i-j2i-jai-jj!(i-j)!(dif(x)dxi|x=a2+ai+jdig(x)dxi|x=a2).
Formula (2.59) becomes(2.62)∑n=0∞an(a2-2abc+c2)n/2=∑i=0∞∑j=0iAijbi-jci+j.


#### 2.3.1. The Case of the Law of Cosines

Consider a numberθ.Let(2.63)b=cos⁡θ.The following formula appears in [2]:(2.64)(cos⁡θ)n=21-nn!∑k=0n/2̲cos⁡((n-2k)θ)k!(n-k)!(1+δ(n,2k)).
Let us use this formula to rewrite formula (2.62):
(2.65)∑n=0∞an(a2-2accos⁡θ+c2)n/2=∑i=0∞∑j=0iAij21-i+j(i-j)!∑k=0i/2-j/2̲cos⁡((i-j-2k)θ)k!(i-j-k)!(1+δ(i-j,2k))ci+j.
By Lemma A.2 (see the appendix), the following formula holds:(2.66)∑i=0∞∑j=0iAij21-i+j(i-j)!∑k=0i/2-j/2̲cos⁡((i-j-2k)θ)k!(i-j-k)!(1+δ(i-j,2k))ci+j=∑i=0∞∑j=0i∑k=0j21-i+jAi+k,j-k4-k(i+2k-j)!cos⁡((i-j)θ)ci+jk!(i+k-j)!(1+δ(i,j)).
By formula (2.61), the following formula holds:(2.67)Ai+k,j-k=(-1)i-j2i+2k-jai+2k-j(di+kf(x)/dxi+k|x=a2+ai+j(di+kg(x)/dxi+k)|x=a2)(j-k)!(i+2k-j)!.
Multiply by 21-i+j4-k(i+2k-j)!/k!(i+k-j)!(1+δ(i,j)):(2.68)21-i+jAi+k,j-k4-k(i+2k-j)!k!(i+k-j)!(1+δ(i,j))=2(-1)i-j(ai+2k-j(di+kf(x)/dxi+k)|x=a2+a2i+2k(di+kg(x)/dxi+k)|x=a2)(j-k)!k!(i+k-j)!(1+δ(i,j)).
Let us use formula (2.34) to rewrite(2.69)2(-1)i-j(ai+2k-j(di+kf(x)/dxi+k)|x=a2+a2i+2k(di+kg(x)/dxi+k)|x=a2)(j-k)!k!(i+k-j)!(1+δ(i,j))=2(-1)i-j(dj-kxi/dxj-k)|x=a2(a-i+j(di+kf(x)/dxi+k)|x=a2+a2j(di+kg(x)/dxi+k)|x=a2)(j-k)!k!(1+δ(i,j))i!.
Let us add with respect to the index k from zero to j:(2.70)∑k=0j2(-1)i-j(dj-kxi/dxj-k)|x=a2(a-i+j(di+kf(x)/dxi+k)|x=a2+a2jdi+kg(x)/dxi+k)|x=a2)(j-k)!k!(1+δ(i,j))i!=2(-1)i-j(a-i+j(dj/dxj)(xi(dif(x)/dxi))|x=a2+a2j(dj/dxj)(xi(dig(x)/dxi))|x=a2)i!j!(1+δ(i,j)).
Multiply by cos((i-j)θ)ci+j.Let us add with respect to the indexj from zero to i.Let us add with respect to the indexi from zero to infinity.By LemmaA.1 (see the appendix), the following formula holds:(2.71)∑i=0∞∑j=0i2(-1)i-ja2j(dj/dxj)(a-i-jxi(dif(x)/dxi)+xi(dig(x)/dxi))|x=a2i!j!(1+δ(i,j))cos⁡((i-j)θ)ci+j=∑j=0∞∑i=0∞2(-1)j(di/dxi(xi+j(di+j/dxi+j)(a-jf(x)+a2ig(x))))cos⁡(jθ)c2i+ji!(i+j)!(1+δ(0,j)).
We will denote (2.72)Bij=2(-1)j(di/dxi)(xi+j(di+j/dxi+j)(a-jf(x)+a2ig(x)))|x=a2i!(i+j)!(1+δ(0,j)).
The left side of (2.65) becomes(2.73)∑n=0∞an(a2-2accos⁡θ+c2)n/2=∑j=0∞∑i=0∞Bijcos⁡(jθ)c2i+j.(1) A Formula for Calculating Coefficients
Consider two natural numbersm and n.
Let(2.74)f(x)=xm+n,g(x)=0.
Formula (2.72) becomes
(2.75)Bij=2(-1)ja-j+2m+2n-2i((m+n)!)2(m+n-i)!(m+n-i-j)!i!(i+j)!(1+δ(0,j)).


## 2.1. Coefficients with Three Indices

Let us calculate the numbersai+l,j-l,l. By formula (2.10), the following formula holds:(2.22)ai+l,j-l,l=∑k=2li+2l-jai+l-k,j-lbkl.
Change the index k for index k-2l:(2.23)∑k=2li+2l-jai+l-k,j-lbkl=∑k=0i-jai-l-k,j-lb2l+k,l.
By formula (2.7), the following formula holds:(2.24)b2l+k,l=(-1)l(2l+2k)!22l+kl!(l+k)!k!.
Let us use the formula(2.25)dndxn1x=(-1)n(2n)!4-nx-n-1/2n!
to write(2.26)(-1)l(2l+2k)!22l+kl!(l+k)!k!=(-1)k2kxl+k+1/2l!k!dl+kdxl+k1x.
We will substitute this in (2.23):(2.27)∑k=0i-jai-l-k,j-lb2l+k,l=∑k=0i-jai-l-k,j-l(-1)k2kxl+k+1/2l!k!dl+kdxl+k1x.
Formula (2.22) becomes(2.28)ai+l,j-l,l=∑k=0i-jai-l-k,j-l(-1)k2kxl+k+1/2l!k!dl+kdxl+k1x.


## 2.2. A Series of Half-Integer Powers

Consider three numbersa, b, and c.We will develop the power of the sum of three numbers.Lemma 2.1.
Consider a natural numbern. Then, the following formula holds:
(2.29)(a+b+c)n=∑i=0n∑j=0in!an-ibi-jcj(n-i)!j!(i-j)!.Consider the series∑n=0∞an(a2-2abc+c2)n.By Lemma2.1 the following formula holds:(2.30)∑n=0∞an(a2-2abc+c2)n=∑n=0∞an∑i=0n∑j=0i(-1)i-j2i-jn!a2n-i-jbi-jci+j(n-i)!j!(i-j)!.
We will distribute this sum, and we will exchange the order of addition:(2.31)∑n=0∞an(a2-2abc+c2)n=∑i=0∞∑j=0i∑n=i∞an(-1)i-j2i-jn!a2n-i-jbi-jci+j(n-i)!j!(i-j)!.
Denote by(2.32)aij=∑n=i∞an(-1)i-j2i-jn!a2n-i-j(n-i)!j!(i-j)!.
The series becomes(2.33)∑n=0∞an(a2-2abc+c2)n=∑i=0∞∑j=0iaijbi-jci+j.
Consider the formula(2.34)dixndxi|x=a2=a2n-2in!(n-i)!.


### 2.2.1. Formulas for Coefficients with Two Indices

Let us use formula (2.34) to rewrite formula (2.32).(2.35)aij=∑n=i∞an(-1)i-j2i-jai-jj!(i-j)!dixndxi|x=a2.
Consider the function(2.36)f(x)=∑n=0∞anxn.
Let us differentiate the series term by term:(2.37)didxif(x)=∑n=0∞andixndxi.
The i-1 first terms in this series are equal to zero. Then, the following formula holds:(2.38)didxif(x)=∑n=i∞andixndxi.
The number aij becomes(2.39)aij=(-1)i-j2i-jai-jj!(i-j)!dif(x)dxi|x=a2.(1) Coefficients with Two Indices
We write the numbersai-l-k,j-l substituting the numbers aij for numbers ai+j-1aij.
By formula (2.39), the following formula holds:
(2.40)ai+j-k-2l-1ai-l-k,j-l=(-1)i-k-j2i-k-ja2i-2k-2l-1(j-l)!(i-k-j)!di-k-lf(x)dxi-k-l|x=a2.
Let us substitute in (2.28):
(2.41)ai+l,j-l,l=∑k=0i-j(-1)i-k-j2i-k-ja2i-2k-2l-1(j-l)!(i-k-j)!di-k-lf(x)dxi-k-l|x=a2(-1)k2kxl+k+1/2l!k!dl+kdxl+k1x.
Let
(2.42)x=a2.
Then, the following formula holds:
(2.43)ai+l,j-l,l=∑k=0i-j(-1)i-j2i-ja2i(j-l)!(i-k-j)!l!k!di-k-lf(x)dxi-k-l|x=a2dl+kdxl+k1x|x=a2.
The sum equals
(2.44)ai+l,j-l,l=(-1)i-j2i-ja2i(i-j)!(j-l)!l!di-jdxi-j(dj-lf(x)dxj-ldldxl1x)|x=a2.
Let us add with respect of index l from zero to j:
(2.45)∑l=0jai+l,j-l,l=∑l=0j(-1)i-j2i-ja2i(i-j)!(j-l)!l!di-jdxi-j(dj-lf(x)dxj-ldldxl1x)|x=a2=(-1)i-j2i-ja2i(i-j)!di-jdxi-j∑l=0j1(j-l)!l!dj-lf(x)dxj-ldldxl1x|x=a2.
The sum equals
(2.46)∑l=0j1(j-l)!l!dj-lf(x)dxj-ldldxl1x=1j!djdxjf(x)x.
We will substitute this sum in (2.45):
(2.47)∑l=0jai+l,j-l,l=(-1)i-j2i-ja2i(i-j)!di-jdxi-j1j!djdxjf(x)x|x=a2=(-1)i-j2i-ja2i(i-j)!j!didxif(x)x|x=a2.
We will substitute this sum in (2.19):
(2.48)Aij=(-1)i-j2i-ja2ij!(i-j)!didxif(x)x|x=a2.


### 2.2.2. A Series of Half-Integer Powers

Divide both sides of (2.33) by a2-2abc+c2:(2.49)∑n=0∞an(a2-2abc+c2)na2-2abc+c2=∑i=0∞∑j=0iaijbi-jci+ja2-2abc+c2=∑i=0∞∑j=0iaijbi-jci+ja1-2(bc/a)+c2/a2.
By formula (2.21), the following formula holds:(2.50)∑i=0∞∑j=0iaijbi-jci+ja1-2(bc/a)+c2/a2=∑i=0∞∑j=0iAijci+jbi-j,
where  (i)
coefficientsaij in (2.21) are replaced by coefficients ai+j-1aij,(ii)
coefficientsAij are given by formula (2.48).The left side of (2.49) becomes(2.51)∑n=0∞an(a2-2abc+c2)na2-2abc+c2=∑i=0∞∑j=0iAijci+jbi-j.


## 2.2.1. Formulas for Coefficients with Two Indices

Let us use formula (2.34) to rewrite formula (2.32).(2.35)aij=∑n=i∞an(-1)i-j2i-jai-jj!(i-j)!dixndxi|x=a2.
Consider the function(2.36)f(x)=∑n=0∞anxn.
Let us differentiate the series term by term:(2.37)didxif(x)=∑n=0∞andixndxi.
The i-1 first terms in this series are equal to zero. Then, the following formula holds:(2.38)didxif(x)=∑n=i∞andixndxi.
The number aij becomes(2.39)aij=(-1)i-j2i-jai-jj!(i-j)!dif(x)dxi|x=a2.(1) Coefficients with Two Indices
We write the numbersai-l-k,j-l substituting the numbers aij for numbers ai+j-1aij.
By formula (2.39), the following formula holds:
(2.40)ai+j-k-2l-1ai-l-k,j-l=(-1)i-k-j2i-k-ja2i-2k-2l-1(j-l)!(i-k-j)!di-k-lf(x)dxi-k-l|x=a2.
Let us substitute in (2.28):
(2.41)ai+l,j-l,l=∑k=0i-j(-1)i-k-j2i-k-ja2i-2k-2l-1(j-l)!(i-k-j)!di-k-lf(x)dxi-k-l|x=a2(-1)k2kxl+k+1/2l!k!dl+kdxl+k1x.
Let
(2.42)x=a2.
Then, the following formula holds:
(2.43)ai+l,j-l,l=∑k=0i-j(-1)i-j2i-ja2i(j-l)!(i-k-j)!l!k!di-k-lf(x)dxi-k-l|x=a2dl+kdxl+k1x|x=a2.
The sum equals
(2.44)ai+l,j-l,l=(-1)i-j2i-ja2i(i-j)!(j-l)!l!di-jdxi-j(dj-lf(x)dxj-ldldxl1x)|x=a2.
Let us add with respect of index l from zero to j:
(2.45)∑l=0jai+l,j-l,l=∑l=0j(-1)i-j2i-ja2i(i-j)!(j-l)!l!di-jdxi-j(dj-lf(x)dxj-ldldxl1x)|x=a2=(-1)i-j2i-ja2i(i-j)!di-jdxi-j∑l=0j1(j-l)!l!dj-lf(x)dxj-ldldxl1x|x=a2.
The sum equals
(2.46)∑l=0j1(j-l)!l!dj-lf(x)dxj-ldldxl1x=1j!djdxjf(x)x.
We will substitute this sum in (2.45):
(2.47)∑l=0jai+l,j-l,l=(-1)i-j2i-ja2i(i-j)!di-jdxi-j1j!djdxjf(x)x|x=a2=(-1)i-j2i-ja2i(i-j)!j!didxif(x)x|x=a2.
We will substitute this sum in (2.19):
(2.48)Aij=(-1)i-j2i-ja2ij!(i-j)!didxif(x)x|x=a2.


## 2.2.2. A Series of Half-Integer Powers

Divide both sides of (2.33) by a2-2abc+c2:(2.49)∑n=0∞an(a2-2abc+c2)na2-2abc+c2=∑i=0∞∑j=0iaijbi-jci+ja2-2abc+c2=∑i=0∞∑j=0iaijbi-jci+ja1-2(bc/a)+c2/a2.
By formula (2.21), the following formula holds:(2.50)∑i=0∞∑j=0iaijbi-jci+ja1-2(bc/a)+c2/a2=∑i=0∞∑j=0iAijci+jbi-j,
where  (i)
coefficientsaij in (2.21) are replaced by coefficients ai+j-1aij,(ii)
coefficientsAij are given by formula (2.48).The left side of (2.49) becomes(2.51)∑n=0∞an(a2-2abc+c2)na2-2abc+c2=∑i=0∞∑j=0iAijci+jbi-j.


## 2.3. A Series of Half-Integer Powers

Consider the series∑n=0∞anxn/2.Let us separate this sum into(i)
a sum with respect to even indices,(ii)
a sum over odd indices.Then, one has(2.52)∑n=0∞anxn/2=∑n=0∞a2nxn+∑n=0∞a1+2nx1/2+n=∑n=0∞a2nxn+∑n=0∞a1+2nxn+1x.
In the second sum, we change the index n for index n′=n+1:(2.53)∑n=0∞a1+2nxn+1=∑n′=1∞a-1+2n′xn′.
Define(2.54)a-1=0.
Then, the following formula holds:(2.55)∑n′=1∞a-1+2n′xn′=∑n=0∞a-1+2nxn.
Formula (2.52) becomes(2.56)∑n=0∞anxn/2=∑n=0∞a2nxn+∑n=0∞a-1+2nxnx.
We will denote  (2.57)f(x)=∑n=0∞a2nxn,g(x)=∑n=0∞a-1+2nxnx.
Let(2.58)x=a2-2abc+c2.
By formulas (2.33) and (2.51), the following formula holds:(2.59)∑n=0∞an(a2-2abc+c2)n/2=∑i=0∞∑j=0iaijbi-jci+j+∑i=0∞∑j=0ibijbi-jci+j,
where (see (2.39) and (2.48)) (2.60)aij=(-1)i-j2i-jai-jj!(i-j)!dif(x)dxi|x=a2,bij=(-1)i-j2i-ja2ij!(i-j)!dig(x)dxi|x=a2.
We will denote(2.61)Aij=(-1)i-j2i-jai-jj!(i-j)!(dif(x)dxi|x=a2+ai+jdig(x)dxi|x=a2).
Formula (2.59) becomes(2.62)∑n=0∞an(a2-2abc+c2)n/2=∑i=0∞∑j=0iAijbi-jci+j.


### 2.3.1. The Case of the Law of Cosines

Consider a numberθ.Let(2.63)b=cos⁡θ.The following formula appears in [2]:(2.64)(cos⁡θ)n=21-nn!∑k=0n/2̲cos⁡((n-2k)θ)k!(n-k)!(1+δ(n,2k)).
Let us use this formula to rewrite formula (2.62):
(2.65)∑n=0∞an(a2-2accos⁡θ+c2)n/2=∑i=0∞∑j=0iAij21-i+j(i-j)!∑k=0i/2-j/2̲cos⁡((i-j-2k)θ)k!(i-j-k)!(1+δ(i-j,2k))ci+j.
By Lemma A.2 (see the appendix), the following formula holds:(2.66)∑i=0∞∑j=0iAij21-i+j(i-j)!∑k=0i/2-j/2̲cos⁡((i-j-2k)θ)k!(i-j-k)!(1+δ(i-j,2k))ci+j=∑i=0∞∑j=0i∑k=0j21-i+jAi+k,j-k4-k(i+2k-j)!cos⁡((i-j)θ)ci+jk!(i+k-j)!(1+δ(i,j)).
By formula (2.61), the following formula holds:(2.67)Ai+k,j-k=(-1)i-j2i+2k-jai+2k-j(di+kf(x)/dxi+k|x=a2+ai+j(di+kg(x)/dxi+k)|x=a2)(j-k)!(i+2k-j)!.
Multiply by 21-i+j4-k(i+2k-j)!/k!(i+k-j)!(1+δ(i,j)):(2.68)21-i+jAi+k,j-k4-k(i+2k-j)!k!(i+k-j)!(1+δ(i,j))=2(-1)i-j(ai+2k-j(di+kf(x)/dxi+k)|x=a2+a2i+2k(di+kg(x)/dxi+k)|x=a2)(j-k)!k!(i+k-j)!(1+δ(i,j)).
Let us use formula (2.34) to rewrite(2.69)2(-1)i-j(ai+2k-j(di+kf(x)/dxi+k)|x=a2+a2i+2k(di+kg(x)/dxi+k)|x=a2)(j-k)!k!(i+k-j)!(1+δ(i,j))=2(-1)i-j(dj-kxi/dxj-k)|x=a2(a-i+j(di+kf(x)/dxi+k)|x=a2+a2j(di+kg(x)/dxi+k)|x=a2)(j-k)!k!(1+δ(i,j))i!.
Let us add with respect to the index k from zero to j:(2.70)∑k=0j2(-1)i-j(dj-kxi/dxj-k)|x=a2(a-i+j(di+kf(x)/dxi+k)|x=a2+a2jdi+kg(x)/dxi+k)|x=a2)(j-k)!k!(1+δ(i,j))i!=2(-1)i-j(a-i+j(dj/dxj)(xi(dif(x)/dxi))|x=a2+a2j(dj/dxj)(xi(dig(x)/dxi))|x=a2)i!j!(1+δ(i,j)).
Multiply by cos((i-j)θ)ci+j.Let us add with respect to the indexj from zero to i.Let us add with respect to the indexi from zero to infinity.By LemmaA.1 (see the appendix), the following formula holds:(2.71)∑i=0∞∑j=0i2(-1)i-ja2j(dj/dxj)(a-i-jxi(dif(x)/dxi)+xi(dig(x)/dxi))|x=a2i!j!(1+δ(i,j))cos⁡((i-j)θ)ci+j=∑j=0∞∑i=0∞2(-1)j(di/dxi(xi+j(di+j/dxi+j)(a-jf(x)+a2ig(x))))cos⁡(jθ)c2i+ji!(i+j)!(1+δ(0,j)).
We will denote (2.72)Bij=2(-1)j(di/dxi)(xi+j(di+j/dxi+j)(a-jf(x)+a2ig(x)))|x=a2i!(i+j)!(1+δ(0,j)).
The left side of (2.65) becomes(2.73)∑n=0∞an(a2-2accos⁡θ+c2)n/2=∑j=0∞∑i=0∞Bijcos⁡(jθ)c2i+j.(1) A Formula for Calculating Coefficients
Consider two natural numbersm and n.
Let(2.74)f(x)=xm+n,g(x)=0.
Formula (2.72) becomes
(2.75)Bij=2(-1)ja-j+2m+2n-2i((m+n)!)2(m+n-i)!(m+n-i-j)!i!(i+j)!(1+δ(0,j)).


## 2.3.1. The Case of the Law of Cosines

Consider a numberθ.Let(2.63)b=cos⁡θ.The following formula appears in [2]:(2.64)(cos⁡θ)n=21-nn!∑k=0n/2̲cos⁡((n-2k)θ)k!(n-k)!(1+δ(n,2k)).
Let us use this formula to rewrite formula (2.62):
(2.65)∑n=0∞an(a2-2accos⁡θ+c2)n/2=∑i=0∞∑j=0iAij21-i+j(i-j)!∑k=0i/2-j/2̲cos⁡((i-j-2k)θ)k!(i-j-k)!(1+δ(i-j,2k))ci+j.
By Lemma A.2 (see the appendix), the following formula holds:(2.66)∑i=0∞∑j=0iAij21-i+j(i-j)!∑k=0i/2-j/2̲cos⁡((i-j-2k)θ)k!(i-j-k)!(1+δ(i-j,2k))ci+j=∑i=0∞∑j=0i∑k=0j21-i+jAi+k,j-k4-k(i+2k-j)!cos⁡((i-j)θ)ci+jk!(i+k-j)!(1+δ(i,j)).
By formula (2.61), the following formula holds:(2.67)Ai+k,j-k=(-1)i-j2i+2k-jai+2k-j(di+kf(x)/dxi+k|x=a2+ai+j(di+kg(x)/dxi+k)|x=a2)(j-k)!(i+2k-j)!.
Multiply by 21-i+j4-k(i+2k-j)!/k!(i+k-j)!(1+δ(i,j)):(2.68)21-i+jAi+k,j-k4-k(i+2k-j)!k!(i+k-j)!(1+δ(i,j))=2(-1)i-j(ai+2k-j(di+kf(x)/dxi+k)|x=a2+a2i+2k(di+kg(x)/dxi+k)|x=a2)(j-k)!k!(i+k-j)!(1+δ(i,j)).
Let us use formula (2.34) to rewrite(2.69)2(-1)i-j(ai+2k-j(di+kf(x)/dxi+k)|x=a2+a2i+2k(di+kg(x)/dxi+k)|x=a2)(j-k)!k!(i+k-j)!(1+δ(i,j))=2(-1)i-j(dj-kxi/dxj-k)|x=a2(a-i+j(di+kf(x)/dxi+k)|x=a2+a2j(di+kg(x)/dxi+k)|x=a2)(j-k)!k!(1+δ(i,j))i!.
Let us add with respect to the index k from zero to j:(2.70)∑k=0j2(-1)i-j(dj-kxi/dxj-k)|x=a2(a-i+j(di+kf(x)/dxi+k)|x=a2+a2jdi+kg(x)/dxi+k)|x=a2)(j-k)!k!(1+δ(i,j))i!=2(-1)i-j(a-i+j(dj/dxj)(xi(dif(x)/dxi))|x=a2+a2j(dj/dxj)(xi(dig(x)/dxi))|x=a2)i!j!(1+δ(i,j)).
Multiply by cos((i-j)θ)ci+j.Let us add with respect to the indexj from zero to i.Let us add with respect to the indexi from zero to infinity.By LemmaA.1 (see the appendix), the following formula holds:(2.71)∑i=0∞∑j=0i2(-1)i-ja2j(dj/dxj)(a-i-jxi(dif(x)/dxi)+xi(dig(x)/dxi))|x=a2i!j!(1+δ(i,j))cos⁡((i-j)θ)ci+j=∑j=0∞∑i=0∞2(-1)j(di/dxi(xi+j(di+j/dxi+j)(a-jf(x)+a2ig(x))))cos⁡(jθ)c2i+ji!(i+j)!(1+δ(0,j)).
We will denote (2.72)Bij=2(-1)j(di/dxi)(xi+j(di+j/dxi+j)(a-jf(x)+a2ig(x)))|x=a2i!(i+j)!(1+δ(0,j)).
The left side of (2.65) becomes(2.73)∑n=0∞an(a2-2accos⁡θ+c2)n/2=∑j=0∞∑i=0∞Bijcos⁡(jθ)c2i+j.(1) A Formula for Calculating Coefficients
Consider two natural numbersm and n.
Let(2.74)f(x)=xm+n,g(x)=0.
Formula (2.72) becomes
(2.75)Bij=2(-1)ja-j+2m+2n-2i((m+n)!)2(m+n-i)!(m+n-i-j)!i!(i+j)!(1+δ(0,j)).


## 3. Functions on the Plane Expressed as a Linear Combination of Powers of Distances to Points

We deduce an identity regarding linear combinations of powers of distances from certain points, as well as similar new identities, from the formula of a power of the cosine. The formula of a power of the cosine is a powerful tool in mathematical analysis. In the space of analytic functions, we construct sums∑i=1Nai|p-pi|p associated to them. We allow ourselves to use the name sum even if it is not finite.Consider(i)
two numbersr and θ,(ii)
two natural numbersm and n.Assume that there are (i) numbersθ1,θ2,…, (ii) positive numbers p1,p2,…, and (iii) series ∑i=0∞aki(pk2-2pkrcos(θ-θk)+r2)i/2, k=1,2,…, such that(3.1)rn+2mcos⁡(nθ)=∑k∑i=0∞aki(pk2-2pkrcos⁡(θ-θk)+r2)i/2.
We will denote  (3.2)fk(x)=∑i=0∞ak,2ixi,gk(x)=∑i=0∞ak,-1+2ixix.
By formula (2.73), the following formula holds:(3.3)∑i=0∞aki(pk2-2pkrcos⁡(θ-θk)+r2)i/2=∑j=0∞∑i=0∞Bkijcos⁡(j(θ-θk))r2i+j,
where the coefficients Bkij are given by formula (2.72): (3.4)Bijk=2(-1)j(di/dxi)(xi+j(di+j/dxi+j)(a-jfk(x)+a2igk(x)))|x=a2i!(i+j)!(1+δ(0,j)).
We use the formula(3.5)cos⁡(θ-ϕ)=cos⁡(θ)cos⁡(ϕ)+sin(θ)sin(ϕ).
Formula (3.1) becomes(3.6)rn+2mcos⁡(nθ)=∑k∑j=0∞∑i=0∞Bkij(cos⁡(jθ)cos⁡(jθk)+sin(jθ)sin(jθk))r2i+j.
The following formulas are sufficient conditions for the solution of the equation above:(3.7)θk=kπn,k=1,2,…,∑kcos⁡(jkπn)Bkij=δ(i,m)δ(j,n),∑ksin(jkπn)Bkij=0,i,j=0,1,….


### 3.1. The Product of a Power of the Distance to the Origin and Cosine

Consider(i)
1+m numbers a0,…,am(ii)
1+m positive numbers p0,…,pm.We will denote(3.8)fkl(x)=(-1)kalxm+n,k=1,…,2n;l=0,…,m.
In Section 2.3.1(1), we computed the numbers Bij. Let us denote (3.9)Bkl,ij=2(-1)k+jalpl-j+2m+2n-2i((m+n)!)2(m+n-i)!(m+n-i-j)!i!(i+j)!(1+δ(0,j)).
Multiply by cos(jkπ/n)(sin(jkπ/n), resp.), and add up over the index k from one to 2n, ∑k=12nBkl,ijcos(jkπ/n) (∑l=12nBkl,ijsin(jkπ/n), resp.).We use the formula(3.10)∑k=12n(-1)ke-1jkπ/n=2nδ(j,n).
The sums ∑k=12nBkl,ijcos(jkπ/n) and ∑k=12nBkl,ijsin(jkπ/n) are equal to(3.11)∑k=12nBkl,ijcos⁡(jkπn)=4nδ(j,n)(-1)jalpl-j+2m+2n-2i((m+n)!)2(m+n-i)!(m+n-i-j)!i!(i+j)!(1+δ(0,j))=4nδ(j,n)(-1)nalpln+2m-2i((m+n)!)2(m+n-i)!(m-i)!i!(i+n)!(1+δ(0,n)),∑k=12nBkl,ijsin(jkπn)=0.
We want the numbers al to satisfy the following equation:(3.12)∑l=0m∑k=12nBkl,ijcos⁡(jkπn)=δ(i,m)δ(j,n),i=0,…,m;j=0,1,….
By formula (3.11), this sum is reduced to(3.13)∑l=0malpln+2m-2i=δ(i,m)(1+δ(0,n))(-1)n(n-1)!m!4(m+n)!,i=0,…,m.
This is a system of linear equations with unknowns a0,…,am. The coefficient matrix for the system is pln+2m-2i; i, l=0,…,m.Assume that the numbersp0,…,pm are different from each other. Then,(3.14)al=(1+δ(0,n))(-1)n(n-1)!m!4(m+n)!pl-n∏i≠lpi2pi2-pl2,l=0,…,m.
Equations (3.7) are valid. Therefore,(3.15)rn+2mcos⁡(nθ)=∑l=0m∑k=12n(-1)lal(pl2-2plrcos⁡(θ-θk)+r2)m+n,
where, as in (3.7),(3.16)θk=kπn,k=1,2….


### 3.2. Example: The Difference between Two Paraboloids

Write(3.17)rcos⁡(θ)=(rcos⁡(θ)2+116+r2)-(-rcos⁡(θ)2+116+r2).
In the coordinates (x, y),(3.18)rcos⁡(θ)=x.Figure 3 represents graphically Identity (3.17).Figure 1Figure 2Figure 3


### 3.3. Example: A Combination of Fourth-Degree Surfaces

Write(3.19)r2cos⁡(2θ)=-(18-12rsin(θ)+r2)2+(18+12rcos⁡(θ)+r2)2-(18+12rsin(θ)+r2)2+(18-12rcos⁡(θ)+r2)2.
In the coordinates (x, y),(3.20)r2cos⁡(2θ)=x2-y2.Figure 4 represents graphically Identity (3.19).Figure 4Figure 5
Representation of the differencep-pkl.Figure 6
Graphics of both sides of formula (3.22), n=2,3. In the series on the right, we add up to m=6.


### 3.4. Example: The Product of a Bessel Function and Cosine

Consider the productrn∑m=0∞(((-1)m/m!(n+m)!)r2m)cosnθ.We will denote(3.21)aklm=14(-1)n+k+m(1+δ(0,n))pl-nn!((m+n)!)-2∏i≠lpi2pi2-pl2.
By formula (3.15),(3.22)rn∑m=0∞(-1)mm!(n+m)!r2mcos⁡nθ=∑k=12n∑m=0∞∑l=0maklm(pl2-2plrcos⁡(θ-θk)+r2)m+n.
We will denote  (3.23)p:pointwithcoordinates(rcos⁡θ,rsinθ),pkl:pointwithcoordinates(plcos⁡θk,plsinθk),|p|=r,yskl(x)=xn∑m=l∞aklmxm.
Rewrite formula (3.22) as(3.24)rn∑m=0∞(-1)mm!(n+m)!r2mcos⁡nθ=∑k=12n∑l=0∞skl(|p-pkl|2).


## 3.1. The Product of a Power of the Distance to the Origin and Cosine

Consider(i)
1+m numbers a0,…,am(ii)
1+m positive numbers p0,…,pm.We will denote(3.8)fkl(x)=(-1)kalxm+n,k=1,…,2n;l=0,…,m.
In Section 2.3.1(1), we computed the numbers Bij. Let us denote (3.9)Bkl,ij=2(-1)k+jalpl-j+2m+2n-2i((m+n)!)2(m+n-i)!(m+n-i-j)!i!(i+j)!(1+δ(0,j)).
Multiply by cos(jkπ/n)(sin(jkπ/n), resp.), and add up over the index k from one to 2n, ∑k=12nBkl,ijcos(jkπ/n) (∑l=12nBkl,ijsin(jkπ/n), resp.).We use the formula(3.10)∑k=12n(-1)ke-1jkπ/n=2nδ(j,n).
The sums ∑k=12nBkl,ijcos(jkπ/n) and ∑k=12nBkl,ijsin(jkπ/n) are equal to(3.11)∑k=12nBkl,ijcos⁡(jkπn)=4nδ(j,n)(-1)jalpl-j+2m+2n-2i((m+n)!)2(m+n-i)!(m+n-i-j)!i!(i+j)!(1+δ(0,j))=4nδ(j,n)(-1)nalpln+2m-2i((m+n)!)2(m+n-i)!(m-i)!i!(i+n)!(1+δ(0,n)),∑k=12nBkl,ijsin(jkπn)=0.
We want the numbers al to satisfy the following equation:(3.12)∑l=0m∑k=12nBkl,ijcos⁡(jkπn)=δ(i,m)δ(j,n),i=0,…,m;j=0,1,….
By formula (3.11), this sum is reduced to(3.13)∑l=0malpln+2m-2i=δ(i,m)(1+δ(0,n))(-1)n(n-1)!m!4(m+n)!,i=0,…,m.
This is a system of linear equations with unknowns a0,…,am. The coefficient matrix for the system is pln+2m-2i; i, l=0,…,m.Assume that the numbersp0,…,pm are different from each other. Then,(3.14)al=(1+δ(0,n))(-1)n(n-1)!m!4(m+n)!pl-n∏i≠lpi2pi2-pl2,l=0,…,m.
Equations (3.7) are valid. Therefore,(3.15)rn+2mcos⁡(nθ)=∑l=0m∑k=12n(-1)lal(pl2-2plrcos⁡(θ-θk)+r2)m+n,
where, as in (3.7),(3.16)θk=kπn,k=1,2….


## 3.2. Example: The Difference between Two Paraboloids

Write(3.17)rcos⁡(θ)=(rcos⁡(θ)2+116+r2)-(-rcos⁡(θ)2+116+r2).
In the coordinates (x, y),(3.18)rcos⁡(θ)=x.Figure 3 represents graphically Identity (3.17).Figure 1Figure 2Figure 3


## 3.3. Example: A Combination of Fourth-Degree Surfaces

Write(3.19)r2cos⁡(2θ)=-(18-12rsin(θ)+r2)2+(18+12rcos⁡(θ)+r2)2-(18+12rsin(θ)+r2)2+(18-12rcos⁡(θ)+r2)2.
In the coordinates (x, y),(3.20)r2cos⁡(2θ)=x2-y2.Figure 4 represents graphically Identity (3.19).Figure 4Figure 5
Representation of the differencep-pkl.Figure 6
Graphics of both sides of formula (3.22), n=2,3. In the series on the right, we add up to m=6.


## 3.4. Example: The Product of a Bessel Function and Cosine

Consider the productrn∑m=0∞(((-1)m/m!(n+m)!)r2m)cosnθ.We will denote(3.21)aklm=14(-1)n+k+m(1+δ(0,n))pl-nn!((m+n)!)-2∏i≠lpi2pi2-pl2.
By formula (3.15),(3.22)rn∑m=0∞(-1)mm!(n+m)!r2mcos⁡nθ=∑k=12n∑m=0∞∑l=0maklm(pl2-2plrcos⁡(θ-θk)+r2)m+n.
We will denote  (3.23)p:pointwithcoordinates(rcos⁡θ,rsinθ),pkl:pointwithcoordinates(plcos⁡θk,plsinθk),|p|=r,yskl(x)=xn∑m=l∞aklmxm.
Rewrite formula (3.22) as(3.24)rn∑m=0∞(-1)mm!(n+m)!r2mcos⁡nθ=∑k=12n∑l=0∞skl(|p-pkl|2).



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*Source: 713241-2011-01-19.xml*