There is a backward differential equation
$$\frac{\partial}{\partial t} F(x,t) + \frac{\sigma^2}{2}\frac{\partial^2}{\partial x^2} F(x,t) = 0$$
for a function F(x,t). I see that the textbook rewrites this equation as an approximate difference on a two-step binomial x,t lattice with all transition probabilities = 1/2.
And it approximates that $$\frac{\partial}{\partial t} F(x,t) \approx \frac{F(x,t+2dt) - F(x,t)}{2dt}$$ I can understand this step. But it also shows that $$\frac{\partial^2 F}{\partial x^2} = \frac{F(x+2dx,t+2dt) - 2F(x,t+2dt) + F(x-2dx.t+2dt)}{4\sigma^2dt}$$ I'm a little confused about how to achieve this step. And I'm curious why it represents a backward equation. The textbook says that if you write down the discrete difference equation that corresponds to the first equation, it will explain why. Could someone explain what this sentence means? Thank you very much.
You can apply the same step to approximate in differences the second derivative from the first derivative, remembering that $$\frac{\partial^2 F}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial F}{\partial x}\right).$$ I mean, you can use several finite differences approximations (forward, backward, or central differences -I recommend reviewing Wikipedia on this topic, https://en.wikipedia.org/wiki/Finite_difference -) to express your second derivative, like this $$\frac{\partial }{\partial x}\left(\frac{\partial F}{\partial x}\right)\approx\frac{{\left(\frac{\partial F}{\partial x}\right)}_{forward} -{\left(\frac{\partial F}{\partial x}\right)}_{backward} }{2\mathrm{dx}}=\frac{\frac{F\left(x+2dx,t\right)-F\left(x,t\right)}{2\mathrm{dx}}-\frac{F\left(x,t\right)-F\left(x-2\mathrm{dx},t\right)}{2\mathrm{dx}}}{2\mathrm{dx}}$$ $$=\frac{F\left(x+2dx,t\right)-2F\left(x,t\right)+F\left(x-2\mathrm{dx},t\right)}{{\left(2\mathrm{dx}\right)}^2 }=\frac{F\left(x+2dx,t\right)-2F\left(x,t\right)+F\left(x-2\mathrm{dx},t\right)}{4\;\sigma^2 \;\mathrm{dt}}.$$ Hope it helps.