Carlson Symmetric Form - Series Expansion

Series Expansion

In obtaining a Taylor series expansion for or it proves convenient to expand about the mean value of the several arguments. So for, letting the mean value of the arguments be, and using homogeneity, define, and by

\begin{align}R_{F}(x,y,z) & = R_{F}(A (1 - \Delta x),A (1 - \Delta y),A (1 - \Delta z)) \\ & = \frac{1}{\sqrt{A}} R_{F}(1 - \Delta x,1 - \Delta y,1 - \Delta z) \end{align}

that is etc. The differences, and are defined with this sign (such that they are subtracted), in order to be in agreement with Carlson's papers. Since is symmetric under permutation of, and, it is also symmetric in the quantities, and . It follows that both the integrand of and its integral can be expressed as functions of the elementary symmetric polynomials in, and which are

Expressing the integrand in terms of these polynomials, performing a multidimensional Taylor expansion and integrating term-by-term...

\begin{align}R_{F}(x,y,z) & = \frac{1}{2 \sqrt{A}} \int _{0}^{\infty}\frac{1}{\sqrt{(t + 1)^{3} - (t + 1)^{2} E_{1} + (t + 1) E_{2} - E_{3}}} dt \\ & = \frac{1}{2 \sqrt{A}} \int _{0}^{\infty}\left( \frac{1}{(t + 1)^{\frac{3}{2}}} - \frac{E_{2}}{2 (t + 1)^{\frac{7}{2}}} + \frac{E_{3}}{2 (t + 1)^{\frac{9}{2}}} + \frac{3 E_{2}^{2}}{8 (t + 1)^{\frac{11}{2}}} - \frac{3 E_{2} E_{3}}{4 (t + 1)^{\frac{13}{2}}} + O(E_{1}) + O(\Delta^{6})\right) dt \\ & = \frac{1}{\sqrt{A}} \left( 1 - \frac{1}{10} E_{2} + \frac{1}{14} E_{3} + \frac{1}{24} E_{2}^{2} - \frac{3}{44} E_{2} E_{3} + O(E_{1}) + O(\Delta^{6})\right) \end{align}

The advantage of expanding about the mean value of the arguments is now apparent; it reduces identically to zero, and so eliminates all terms involving - which otherwise would be the most numerous.

An ascending series for may be found in a similar way. There is a slight difficulty because is not fully symmetric; its dependence on its fourth argument, is different from its dependence on, and . This is overcome by treating as a fully symmetric function of five arguments, two of which happen to have the same value . The mean value of the arguments is therefore take to be

and the differences, and defined by

\begin{align}R_{J}(x,y,z,p) & = R_{J}(A (1 - \Delta x),A (1 - \Delta y),A (1 - \Delta z),A (1 - \Delta p)) \\ & = \frac{1}{A^{\frac{3}{2}}} R_{J}(1 - \Delta x,1 - \Delta y,1 - \Delta z,1 - \Delta p) \end{align}

The elementary symmetric polynomials in, and (again) are in full

However, it is possible to simplify the formulae for, and using the fact that . Expressing the integrand in terms of these polynomials, performing a multidimensional Taylor expansion and integrating term-by-term as before...

\begin{align}R_{J}(x,y,z,p) & = \frac{3}{2 A^{\frac{3}{2}}} \int _{0}^{\infty}\frac{1}{\sqrt{(t + 1)^{5} - (t + 1)^{4} E_{1} + (t + 1)^{3} E_{2} - (t + 1)^{2} E_{3} + (t + 1) E_{4} - E_{5}}} dt \\ & = \frac{3}{2 A^{\frac{3}{2}}} \int _{0}^{\infty}\left( \frac{1}{(t + 1)^{\frac{5}{2}}} - \frac{E_{2}}{2 (t + 1)^{\frac{9}{2}}} + \frac{E_{3}}{2 (t + 1)^{\frac{11}{2}}} + \frac{3 E_{2}^{2} - 4 E_{4}}{8 (t + 1)^{\frac{13}{2}}} + \frac{2 E_{5} - 3 E_{2} E_{3}}{4 (t + 1)^{\frac{15}{2}}} + O(E_{1}) + O(\Delta^{6})\right) dt \\ & = \frac{1}{A^{\frac{3}{2}}} \left( 1 - \frac{3}{14} E_{2} + \frac{1}{6} E_{3} + \frac{9}{88} E_{2}^{2} - \frac{3}{22} E_{4} - \frac{9}{52} E_{2} E_{3} + \frac{3}{26} E_{5} + O(E_{1}) + O(\Delta^{6})\right) \end{align}

As with, by expanding about the mean value of the arguments, more than half the terms (those involving ) are eliminated.

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