345] ON THE INFLEXIONS OF THE CUBICAL DIVERGENT PARABOLAS.
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Consider next the case of a curve without singularities; and first the complex case,
the condition for which is that the equation ж 2 + 3bx + Зс = 0 may have its roots real,
or c<§6 2 . The values of ж which give y = 0 are
ж = 0, x = — |6 + V3 (§ b 2 — c) ;
and we may without loss of generality assume that b and c are each of them positive ;
the value x = 0 xvill then belong to the vertex of the parabolic portion, and the two
negative values x = — f6 + V3 (fb 2 — c) will belong to the vertices of the oval. The
limiting values c = () and c = § b 2 give the acnodal and the crunodal curves respectively,
which have been already considered.
In the case in question (b = +, c — +, c < f b 2 ), the equation x A + 4bx 3 + беж — 3c 2 = 0
has only two real roots, one of them positive and the other negative ; and the positive
root substituted in the equation у 2 — x (ж 2 + Sbx + 3c) gives y 2 = +, and we have thus
the two real inflexions: in order to verify that the negative root gives imaginary
inflexions, it must be shown that this negative root does not lie between the two
values x = — f b + v3 (f b 2 — c), or, what is the same thing, that these values substituted
for ж in the function
ж 4 + 4 bx 3 + Sex 2 — 3c 2 ,
give results of the same sign.
To verify this write c = f5 2 (l—d 2 ) (where 6< 1) and ж = |6(|—1); then for the
limiting values of x, we have §• b (£ — 1) = — f b + '^Ъв, or £= + #; and moreover
ж 4 + 46ж 3 + беж 2 - Зс 2 = {3 (f - l) 4 + 8 (£ - l) 3 + 6 (£ - l) 2 (1 - в 2 ) - (1 - <9 2 ) 2 },
where the term in brackets is
= 3| 3 - 4£ - 60 2 f 2 + 12d 2 £ - 40 2 - e\
and writing ^ = ± в, this becomes
- 4<9 4 - 4d 2 ± 8d 3 , =-4d 2 (^ + l) 2 ,
so that the two values are each negative, and the theorem is thus proved. It may be
added that the curve
у = ж 4 + 4 bx 3 + беж 2 — Зс 2
cuts the axis in two real points, one of them situate between the oval and the para
bolic portion of the cubic parabola, the other within the parabolic portion.
Lastly, for the simplex case, the condition for which is c>f& 2 ; the equation
0 = ж 4 + 4bx? + беж 2 — 3c 2 has, as before, two real roots, one positive and the other
negative; and since the negative root substituted for ж in the equation у 2 =ж 3 + 36ж 2 + 3сж
gives a negative value of y 2 , it is only the positive root which gives an ordinate
through two real inflexions. The curve у = ж 4 + 46ж 3 + беж 2 — Зс 2 meets the axis in two
real points, one of them without, the other within the cubic parabola.
