I'm reading the notes here, particularly page 33, which gives Einstein's formulation of Brownian motion. One statement made in the construction seems to lack justification.
- Suppose that the probability density of the event that an ink particle moves from $x$ to $x+y$ in (small) time $\tau$ is $f(τ, y)$. Then $$u(x,t+\tau) = \int_{-\infty}^\infty u(x-y,t)f(y,\tau)dy = \int_{-\infty}^\infty \left( u-u_xy+\frac12 u_{xx}y^2+\cdots \right)f(y,\tau)dy.$$
Since $f$ is a probability density, $\int f dy=1$. And since $f$ is symmetric, $\int yfdy=0$. Now if we assume that $\int y^2 fdy=D\tau$, for some constant $D>0$, the author correctly concludes that
$$\frac{u(x,t+\tau)-u(x,t)}{\tau} = \frac D2u_{xx}(x,t) + \{\text{higher order terms}\}.$$
He then says that if $\tau\to0$, the higher order terms vanish and we have $u_t = \frac D2 u_{xx}.$
Question: why do the higher order terms vanish?