Editing Carlson symmetric form (section)

==Series Expansion==

In obtaining a [[Taylor series]] expansion for <math>R_{F}</math> or <math>R_{J}</math> it proves convenient to expand about the mean value of the several arguments. So for <math>R_{F}</math>, letting the mean value of the arguments be <math>A = (x + y + z)/3</math>, and using homogeneity, define <math>\Delta x</math>, <math>\Delta y</math> and <math>\Delta z</math> by

:<math>\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}</math>

that is <math>\Delta x = 1 - x/A</math> etc. The differences <math>\Delta x</math>, <math>\Delta y</math> and <math>\Delta z</math> are defined with this sign (such that they are ''subtracted''), in order to be in agreement with Carlson's papers. Since <math>R_{F}(x,y,z)</math> is symmetric under permutation of <math>x</math>, <math>y</math> and <math>z</math>, it is also symmetric in the quantities <math>\Delta x</math>, <math>\Delta y</math> and <math>\Delta z</math>. It follows that both the integrand of <math>R_{F}</math> and its integral can be expressed as functions of the [[elementary symmetric polynomial]]s in <math>\Delta x</math>, <math>\Delta y</math> and <math>\Delta z</math> which are

:<math>E_{1} = \Delta x + \Delta y + \Delta z = 0</math>

:<math>E_{2} = \Delta x \Delta y + \Delta y \Delta z + \Delta z \Delta x</math>

:<math>E_{3} = \Delta x \Delta y \Delta z</math>

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

:<math>\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}</math>

The advantage of expanding about the mean value of the arguments is now apparent; it reduces <math>E_{1}</math> identically to zero, and so eliminates all terms involving <math>E_{1}</math> - which otherwise would be the most numerous.

An ascending series for <math>R_{J}</math> may be found in a similar way. There is a slight difficulty because <math>R_{J}</math> is not fully symmetric; its dependence on its fourth argument, <math>p</math>, is different from its dependence on <math>x</math>, <math>y</math> and <math>z</math>. This is overcome by treating <math>R_{J}</math> as a fully symmetric function of ''five'' arguments, two of which happen to have the same value <math>p</math>. The mean value of the arguments is therefore taken to be

:<math>A = \frac{x + y + z + 2 p}{5}</math>

and the differences <math>\Delta x</math>, <math>\Delta y</math> <math>\Delta z</math> and <math>\Delta p</math> defined by

:<math>\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}</math>

The [[elementary symmetric polynomial]]s in <math>\Delta x</math>, <math>\Delta y</math>, <math>\Delta z</math>, <math>\Delta p</math> and (again) <math>\Delta p</math> are in full

:<math>E_{1} = \Delta x + \Delta y + \Delta z + 2 \Delta p = 0</math>

:<math>E_{2} = \Delta x \Delta y + \Delta y \Delta z + 2 \Delta z \Delta p + \Delta p^{2} + 2 \Delta p \Delta x + \Delta x \Delta z + 2 \Delta y \Delta p</math>

:<math>E_{3} = \Delta z \Delta p^{2} + \Delta x \Delta p^{2} + 2 \Delta x \Delta y \Delta p + \Delta x \Delta y \Delta z + 2 \Delta y \Delta z \Delta p + \Delta y \Delta p^{2} + 2 \Delta x \Delta z \Delta p</math>

:<math>E_{4} = \Delta y \Delta z \Delta p^{2} + \Delta x \Delta z \Delta p^{2} + \Delta x \Delta y \Delta p^{2} + 2 \Delta x \Delta y \Delta z \Delta p</math>

:<math>E_{5} = \Delta x \Delta y \Delta z \Delta p^{2}</math>

However, it is possible to simplify the formulae for <math>E_{2}</math>, <math>E_{3}</math> and <math>E_{4}</math> using the fact that <math>E_{1} = 0</math>. Expressing the integrand in terms of these polynomials, performing a multidimensional Taylor expansion and integrating term-by-term as before...

:<math>\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}</math>

As with <math>R_{J}</math>, by expanding about the mean value of the arguments, more than half the terms (those involving <math>E_{1}</math>) are eliminated.