Weierstrass Points on Quartics Let $V$ be a smooth projective plane quartic. We say that $P \in V$ is a Weierstrass point if $\dim(L(gP)) \geq 2$. We say that a point $Q$ is a point of inflexion if the tangent line at $Q$ intersects $Q$ with multiplicity at least 3. How do I show that, for the special case of $V$ a plane quartic, these definitions coincide, in the sense that being an inflexion point is equivalent to being a Weierstrass point? Please don't give an answer involving things like schemes and sheaves - let's keep it in the language of varieties. I know the statement of the Riemann-Roch theorem.

A Weierstrass point of a smooth projective curve $C$ (compact Riemann surface, if you prefer) is a ramification point of the complete canonical series $K_C$. In other words, it is a point $P\in C$ such that $h^0(K_C(-gP))>0$, where $g=g_C$. This is equivalent to $h^0(gP)>1$, as you mention (by Riemann-Roch $+$ Serre duality).

> In other words, $P\in C$ is a Weierstrass point iff the canonical series $K_C$ has a (nonzero) section vanishing at least $g_C$ times at $P$.

If $C$ is a smooth plane quartic, any canonical divisor consists of four aligned points on the curve. In other words, the canonical series is the linear series cut out by lines, $$K_C=\mathscr O_C(1).$$ Since $g=3$, a point $P\in C$ is Weierstrass iff $\mathscr O_C(1)$ has a (nonzero) section vanishing at least $3$ times at $P$. But this is the definition of an inflexion point.

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_Aside_. It is true for every smooth plane curve $C$ of degree $d$ that $K_C=\mathscr O_C(d-3)$.