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Daniele Tampieri
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In Simon's book Harmonic Analysis, example 3.5.12 shows:

Fix $x \in \mathbb{R}^{\nu}$ (with $\left.\nu \geq 3\right), x \neq 0$ and define $f$ on $\{y:| y|<| x |\}$ by $$ f(y)=|x-y|^{-(\nu-2)}. $$ Then we have$,$ $$ |x-y|^{-(\nu-2)}=\sum_{d=0}^{\infty} \frac{|y|^{d}}{|x|^{d+\nu-2}} \alpha_{\nu, d} Z_{d}\left(\frac{y}{|y|}, \frac{x}{|x|}\right) \tag{$\ast$} $$$$ |x-y|^{-(\nu-2)}=\sum_{d=0}^{\infty} \frac{|y|^{d}}{|x|^{d+\nu-2}} \alpha_{\nu, d} Z_{d}\left(\frac{y}{|y|}, \frac{x}{|x|}\right) \label{1}\tag{$\ast$} $$ where $\alpha_{\nu, d}$ is a constant which is dependent on $\nu, d$ and the $Z_{d}$ (the zonal harmonic of degree d) term at $y=0$ is interpreted as 1 if $d=0$ and is irrelevant if $d \geq 1$.

My Question is$,$ for fixed $x \in \mathbb{R}^{\nu}$ (with $\left.\nu \geq 3\right), x \neq 0$ and function $f(y)=|x-y|^{-\lambda}$ on $\{y|| y|<| x |\}$ with $\lambda>0,$ can we obtain a similar formula like $(\ast)$\eqref{1} $?$

Any reference and suggestion will be greatly appreciated. Thanks!

In Simon's book Harmonic Analysis, example 3.5.12 shows:

Fix $x \in \mathbb{R}^{\nu}$ (with $\left.\nu \geq 3\right), x \neq 0$ and define $f$ on $\{y:| y|<| x |\}$ by $$ f(y)=|x-y|^{-(\nu-2)}. $$ Then we have$,$ $$ |x-y|^{-(\nu-2)}=\sum_{d=0}^{\infty} \frac{|y|^{d}}{|x|^{d+\nu-2}} \alpha_{\nu, d} Z_{d}\left(\frac{y}{|y|}, \frac{x}{|x|}\right) \tag{$\ast$} $$ where $\alpha_{\nu, d}$ is a constant which is dependent on $\nu, d$ and the $Z_{d}$ (the zonal harmonic of degree d) term at $y=0$ is interpreted as 1 if $d=0$ and is irrelevant if $d \geq 1$.

My Question is$,$ for fixed $x \in \mathbb{R}^{\nu}$ (with $\left.\nu \geq 3\right), x \neq 0$ and function $f(y)=|x-y|^{-\lambda}$ on $\{y|| y|<| x |\}$ with $\lambda>0,$ can we obtain a similar formula like $(\ast)$ $?$

Any reference and suggestion will be greatly appreciated. Thanks!

In Simon's book Harmonic Analysis, example 3.5.12 shows:

Fix $x \in \mathbb{R}^{\nu}$ (with $\left.\nu \geq 3\right), x \neq 0$ and define $f$ on $\{y:| y|<| x |\}$ by $$ f(y)=|x-y|^{-(\nu-2)}. $$ Then we have$,$ $$ |x-y|^{-(\nu-2)}=\sum_{d=0}^{\infty} \frac{|y|^{d}}{|x|^{d+\nu-2}} \alpha_{\nu, d} Z_{d}\left(\frac{y}{|y|}, \frac{x}{|x|}\right) \label{1}\tag{$\ast$} $$ where $\alpha_{\nu, d}$ is a constant which is dependent on $\nu, d$ and the $Z_{d}$ (the zonal harmonic of degree d) term at $y=0$ is interpreted as 1 if $d=0$ and is irrelevant if $d \geq 1$.

My Question is$,$ for fixed $x \in \mathbb{R}^{\nu}$ (with $\left.\nu \geq 3\right), x \neq 0$ and function $f(y)=|x-y|^{-\lambda}$ on $\{y|| y|<| x |\}$ with $\lambda>0,$ can we obtain a similar formula like \eqref{1} $?$

Any reference and suggestion will be greatly appreciated. Thanks!

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Multipole Expansionexpansion

In $\textit{Simon's book Harmonic Analysis},$ $\textbf{example 3.5.12}$ showSimon's book Harmonic Analysis, example 3.5.12 shows:

Fix $x \in \mathbb{R}^{\nu}$ (with $\left.\nu \geq 3\right), x \neq 0$ and define $f$ on $\{y|| y|<| x |\}$$\{y:| y|<| x |\}$ by $$ f(y)=|x-y|^{-(\nu-2)}. $$ Then we have$,$ $$ |x-y|^{-(\nu-2)}=\sum_{d=0}^{\infty} \frac{|y|^{d}}{|x|^{d+\nu-2}} \alpha_{\nu, d} Z_{d}\left(\frac{y}{|y|}, \frac{x}{|x|}\right) \tag{$\ast$} $$ where $\alpha_{\nu, d}$ is a constant which is dependent on $\nu, d$ and the $Z_{d}$ (the zonal harmonic of degree d) term at $y=0$ is interpreted as 1 if $d=0$ and is irrelevant if $d \geq 1$.

$\textbf{My Question}$My Question is$,$ for fixdfixed $x \in \mathbb{R}^{\nu}$ (with $\left.\nu \geq 3\right), x \neq 0$ and function $f(y)=|x-y|^{-\lambda}$ on $\{y|| y|<| x |\}$ with $\lambda>0,$ can we obtain a similar formula like $(\ast)$ $?$

Any reference and suggestion will be greatly appreciated. Thanks!

Multipole Expansion

In $\textit{Simon's book Harmonic Analysis},$ $\textbf{example 3.5.12}$ show :

Fix $x \in \mathbb{R}^{\nu}$ (with $\left.\nu \geq 3\right), x \neq 0$ and define $f$ on $\{y|| y|<| x |\}$ by $$ f(y)=|x-y|^{-(\nu-2)}. $$ Then we have$,$ $$ |x-y|^{-(\nu-2)}=\sum_{d=0}^{\infty} \frac{|y|^{d}}{|x|^{d+\nu-2}} \alpha_{\nu, d} Z_{d}\left(\frac{y}{|y|}, \frac{x}{|x|}\right) \tag{$\ast$} $$ where $\alpha_{\nu, d}$ is a constant which is dependent on $\nu, d$ and the $Z_{d}$ (the zonal harmonic of degree d) term at $y=0$ is interpreted as 1 if $d=0$ and is irrelevant if $d \geq 1$.

$\textbf{My Question}$ is$,$ for fixd $x \in \mathbb{R}^{\nu}$ (with $\left.\nu \geq 3\right), x \neq 0$ and function $f(y)=|x-y|^{-\lambda}$ on $\{y|| y|<| x |\}$ with $\lambda>0,$ can we obtain a similar formula like $(\ast)$ $?$

Any reference and suggestion will be greatly appreciated. Thanks!

Multipole expansion

In Simon's book Harmonic Analysis, example 3.5.12 shows:

Fix $x \in \mathbb{R}^{\nu}$ (with $\left.\nu \geq 3\right), x \neq 0$ and define $f$ on $\{y:| y|<| x |\}$ by $$ f(y)=|x-y|^{-(\nu-2)}. $$ Then we have$,$ $$ |x-y|^{-(\nu-2)}=\sum_{d=0}^{\infty} \frac{|y|^{d}}{|x|^{d+\nu-2}} \alpha_{\nu, d} Z_{d}\left(\frac{y}{|y|}, \frac{x}{|x|}\right) \tag{$\ast$} $$ where $\alpha_{\nu, d}$ is a constant which is dependent on $\nu, d$ and the $Z_{d}$ (the zonal harmonic of degree d) term at $y=0$ is interpreted as 1 if $d=0$ and is irrelevant if $d \geq 1$.

My Question is$,$ for fixed $x \in \mathbb{R}^{\nu}$ (with $\left.\nu \geq 3\right), x \neq 0$ and function $f(y)=|x-y|^{-\lambda}$ on $\{y|| y|<| x |\}$ with $\lambda>0,$ can we obtain a similar formula like $(\ast)$ $?$

Any reference and suggestion will be greatly appreciated. Thanks!

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In $\textit{Simon's book Harmonic Analysis},$ $\textbf{example 3.5.12}$ show :

Fix $x \in \mathbb{R}^{\nu}$ (with $\left.\nu \geq 3\right), x \neq 0$ and define $f$ on $\{y|| y|<| x |\}$ by $$ f(y)=|x-y|^{-(\nu-2)}. $$ Then we have$,$ $$ |x-y|^{-(\nu-2)}=\sum_{d=0}^{\infty} \frac{|y|^{d}}{|x|^{d+\nu-2}} \alpha_{\nu, d} Z_{d}\left(\frac{y}{|y|}, \frac{x}{|x|}\right) \tag{$\ast$} $$ where $\alpha_{\nu, d}$ is a constant which is dependent on $\nu, d$ and the $Z_{d}$ (the zonal harmonic of degree d) term at $y=0$ is interpreted as 1 if $d=0$ and is irrelevant if $d \geq 1$.

$\textbf{My Question}$ is$,$ for fixd $x \in \mathbb{R}^{\nu}$ (with $\left.\nu \geq 3\right), x \neq 0$ and function $f(y)=|x-y|^{-\lambda}$ on $\{y|| y|<| x |\}$ with $\lambda>0,$ can we obtain a similar formula aslike $(\ast)$ $?$

Any reference and suggestion will be greatly appreciated. Thanks!

In $\textit{Simon's book Harmonic Analysis},$ $\textbf{example 3.5.12}$ show :

Fix $x \in \mathbb{R}^{\nu}$ (with $\left.\nu \geq 3\right), x \neq 0$ and define $f$ on $\{y|| y|<| x |\}$ by $$ f(y)=|x-y|^{-(\nu-2)}. $$ Then we have$,$ $$ |x-y|^{-(\nu-2)}=\sum_{d=0}^{\infty} \frac{|y|^{d}}{|x|^{d+\nu-2}} \alpha_{\nu, d} Z_{d}\left(\frac{y}{|y|}, \frac{x}{|x|}\right) \tag{$\ast$} $$ where $\alpha_{\nu, d}$ is a constant which is dependent on $\nu, d$ and the $Z_{d}$ (the zonal harmonic of degree d) term at $y=0$ is interpreted as 1 if $d=0$ and is irrelevant if $d \geq 1$.

$\textbf{My Question}$ is$,$ for fixd $x \in \mathbb{R}^{\nu}$ (with $\left.\nu \geq 3\right), x \neq 0$ and function $f(y)=|x-y|^{-\lambda}$ on $\{y|| y|<| x |\}$ with $\lambda>0,$ can we obtain a similar formula as $(\ast)$ $?$

Any reference and suggestion will be greatly appreciated. Thanks!

In $\textit{Simon's book Harmonic Analysis},$ $\textbf{example 3.5.12}$ show :

Fix $x \in \mathbb{R}^{\nu}$ (with $\left.\nu \geq 3\right), x \neq 0$ and define $f$ on $\{y|| y|<| x |\}$ by $$ f(y)=|x-y|^{-(\nu-2)}. $$ Then we have$,$ $$ |x-y|^{-(\nu-2)}=\sum_{d=0}^{\infty} \frac{|y|^{d}}{|x|^{d+\nu-2}} \alpha_{\nu, d} Z_{d}\left(\frac{y}{|y|}, \frac{x}{|x|}\right) \tag{$\ast$} $$ where $\alpha_{\nu, d}$ is a constant which is dependent on $\nu, d$ and the $Z_{d}$ (the zonal harmonic of degree d) term at $y=0$ is interpreted as 1 if $d=0$ and is irrelevant if $d \geq 1$.

$\textbf{My Question}$ is$,$ for fixd $x \in \mathbb{R}^{\nu}$ (with $\left.\nu \geq 3\right), x \neq 0$ and function $f(y)=|x-y|^{-\lambda}$ on $\{y|| y|<| x |\}$ with $\lambda>0,$ can we obtain a similar formula like $(\ast)$ $?$

Any reference and suggestion will be greatly appreciated. Thanks!

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