曲线的二重切线数$\frac{n}{2}(n-2)(n^2-9)$

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$n$次曲线的最多的二重切线数$=\frac{n}{2}(n-2)(n^2-9)$，如何证明

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github.com/jmokland/schubert/blob/master/schu … t/src/schubertmanual

proj(2,h,all):             # the dual projective plane
proj(2,j,all):             # the projective plane
bundlesection(C,o(d*j)):   # define a plane curve of degree d
morphism(f,C,Ph,[(d-1)*j]):# the Gauss map to the dual plane
multiplepoint(f,2)/2:      # double points are bitangents or flexes
                           # correct for flexes.  Flexes can of 
                           # course be calculated automatically.
bitangents:=expand(integral(C,%)-(3*d*(d-2)));
                                   4    3        2
                bitangents := 1/2 d  - d  - 9/2 d  + 9 d

subs(d=4,%);
                                   28

4次曲线最多有28条二重切线