I am learning about Taylor series to calculate the truncation error of numerical methods.
My problem is the following:
I want to use the Taylor series for $$ f(t,y(t)) = f(t+\frac{h}{2}, y(t)+f\cdot \frac{h}{2}).$$ For the first order I get $\tau = f + \frac{h}{2}f_{,t}+\frac{h}{2}f_{,y}f+\mathrm{O} (h^{2})$
My problem is the 2nd order. Calculating Taylor for $\frac{d^2}{dt^2} f(t,y(t))$ gives me $ f_{,tt} \cdot \frac{h^2}{8}$.
For calulating the Taylor for $\frac{d^2}{dy^2} f(t,y(t))$ I would use the chain rule and the product rule with $u = f_{,y} , u' = f_{,yy}\cdot f, v = f, v' = f_{,y}$. This leads me to $\frac{d^2}{dy^2} f(t,y(t)) = f_{,yy}\cdot f^2 + f_{,y}^2\cdot f $.
BUT: That is not correct, the solutions says it's $\frac{d^2}{dy^2} f(t,y(t)) = f_{,yy}\cdot f^2$. Why is the second part missing? Do I miscalculate? Thanks in advance!
Your result would be correct, after some massive corrections of the notation, for calculating $y'''$. But you are calculating the Taylor series $$ f(t+a, y+b)=f(t,y)+f_{,t}a+f_{,y}b+\frac12f_{,tt}a^2+f_{,ty}ab+\frac12f_{,yy}b^2+... $$
In $f(t,y(t))$, $y$ is a dependent variable. Thus it makes no sense to want to compute the total derivative $\frac{d}{dy}$. You can compute the partial derivative for the second position, that is then $\frac{\partial^2}{\partial y^2}f=f_{,yy}$, without any additional factors.