It would become much easier to figure out the proof of a formula [Hob, p.26,
l.-9] if we
were to consider the formula and its conjugate counterpart [Hob, .26, l.-1]
simultaneously. (Note. The conjugate of a
complex number is a special case of symmetry.)
The symmetric feature of the conditions for various values of the integral
[Hob, p.360, l.14] emerges if we separate each coefficient's real
part from its imaginary part [Hob, p.361, l.-9].
Example. Compare [Hob, p.361, l.13-l.-12] with [Hob,
p.363, l.-5-p.364, l.5].
We may trace the reason why a theorem is symmetric by using the symmetry of
Let t and t0 be variables of a theorem. If we
interchange t and t0 and the theorem remains
the same except for the notational difference, then we say that the theorem is
symmetric with respect to t and t0 or that t
and t0 are symmetric respect to the theorem.
Example. [Har, p.54, (4.2)] is symmetric with respect to t and t0.
Let A=(t0£t) and B=(|y(t)|³u0(t)).
If we interchange t and t0 in (AÞB),
we obtain (A'ÞB'), where A'=(t£t0)
The proof of [Har, p.54, (4.2)] uses (A'ÞB') [Har,
p.55, l.3], which is the symmetric counterpart of the conditional statement (AÞB).
The symmetry between two parameters in coefficients implies the symmetry between
Suppose Ai(z, a, a')
= Ai(z, a', a) (i
= 0, 1, 2).
If u(z, a) is a solution of A2(z, a, a')(d2u/dz2) +
A1(z, a, a')(du/dz) +
A0(z, a, a')u = f(z), so is u(z, a').
Example: [Wat1, p.208, l.-16-l.-11]
The symmetry between exponents for Riemann's P-equation: [Wat1, p.283, l.-6-p.284, l.6]
The symmetry among regular singular points for
Riemann's P-equation: [Wat1, p.284, l.7-p.285, l.-10]
By symmetrizing the hypergeometric equation given in [Wat1, p.207, Example], we obtain
the Riemann's P-equation [Wat1, p.206, l.-13-l.-11]. Similarly, by symmetrizing the
15 contiguous relations given in , we obtain the 30 contiguous relations given in [Wat1,
Remark. Symmetrization may generalize results, but may blur essential ideas.¬