Jackson - p. 35 - taylor expansion of charge density

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SUMMARY

The discussion centers on the Taylor series expansion of charge density, specifically the expression for ρ({\bf{x'}}) around {\bf{x'}} = {\bf{x}}. The correct Taylor expansion is given as ρ({\bf{x'}}) = ρ({\bf{x}}) + (1/6)r²∇²ρ + ..., where the user questions the coefficient of 1/6 instead of 1/2. The consensus confirms that the 1/6 coefficient is accurate, aligning with the standard formulation in physics for second-order terms in Taylor expansions.

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bjnartowt
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Homework Statement



With a Taylor series expansion of the well-behaved [itex]\rho ({\bf{x'}})[/itex] around [itex]{\bf{x'}} = {\bf{x}}[/itex], one finds the Taylor expansion of the charge density to be,

[itex]\rho ({\bf{x'}}) = \rho ({\bf{x}}) + {\textstyle{1 \over 6}}{r^2}{\nabla ^2}\rho + ...[/itex]

Homework Equations





The Attempt at a Solution



This proposed Taylor exapnsion looks correct, except for the 1/6 instead of the 1/2. What gives??
 
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bjnartowt said:

Homework Statement



With a Taylor series expansion of the well-behaved [itex]\rho ({\bf{x'}})[/itex] around [itex]{\bf{x'}} = {\bf{x}}[/itex], one finds the Taylor expansion of the charge density to be,

[itex]\rho ({\bf{x'}}) = \rho ({\bf{x}}) + {\textstyle{1 \over 6}}{r^2}{\nabla ^2}\rho + ...[/itex]


The Attempt at a Solution



This proposed Taylor exapnsion looks correct, except for the 1/6 instead of the 1/2. What gives??

Take a look at this thread.
 

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