Implicit Partial derivative computation for 3rd order Runge Kutta derivation?

I need to derive the 3rd order Runge Kutta method which needs a tedious computation of partial derivatives, which i have a feeling i will make a mistake on eventually. I was wondering if there is any software or something those lines that will help me do this? Here is the setup

$$f = f(t,x)=x'=x'(t)=\frac{dx(t)}{dt}$$ $$x(t+h)=x(t)+hx'(t)+\frac{h^2}{2}x''(t)+\frac{h^3}{6}x'''(t)+O(h^4)$$ $$x(t+h)=x(t)+hf+\frac{h^2}{2}(f_t+f_xf)+\frac{h^3}{6}(f_{tt}+f_{tx}f+f_{xt}f+f_tf_x+f_{xx}f^2+(f_x)^2f)+O(h^4)$$ 3rd order $$O(h^4)=0$$ This is where i need help. It is defined that $$f(x+h,t+k)=\sum\limits_{i=1}^\infty {\frac{1}{i!}(h\frac{\partial}{\partial t}+k\frac{\partial}{\partial x})^i}f(x,t)$$ For the Runge Kutta method, k is a function of x and t. I need to show that $$x(t+h)=x(t)+\frac{1}{9}(2F_1+3F_2+4F_3)$$ Where $$F_1=hf(t,x) , F_2=hf(t+\frac{1}{2}h,x+\frac{1}{2}F_1),F_3=hf(t+\frac{3}{4}h,x+\frac{3}{4}F_2)$$ I need either a detailed explanation on how to derive this or at least some software that can expand $$f(x+ah,t+bk) , k=F(x,t)$$ Thanks

• Thanks LutzL, i'm aware of Butcher tableaux, but the point of this exercise is to not quote the result. I still want a way to expand that formula i posted in a safe way because i know for a fact i'll screw up the derivation – Baklava Gain Mar 17 '14 at 23:12

First, lose the $t$. You can always make it an extra coordinate by setting $\frac{d}{dt}t=1$. Now you only have to specify everything for an autonomous system $x'=f(x)$. There should be derivative tensors involved,... but the nice thing is, the coefficients of the Runge-Kutta methods do not depend on dimensions, so you can just examine the case of dimension 1. As you did in your formulas anyway.
Leaving the leading $h$ out, the following terms need only be examined in $O(h^3)$ since the final result is a combination of $hk_j$ \begin{align} k_1&=f(x)\\ k_2&=f(x+a_{21}hk_1)&&=f+hf'\,a_{21}f+\tfrac12h^2f''\,(a_{21}f)^2\\ k_3&=f(x+a_{32}hk_2)&&=f+hf'\,a_{32}(f+hf'\,a_{21}f)+\tfrac12h^2f''\,(a_{32}f)^2\\[0.4em]\hline x_+&=x+h(b_1k_1+b_2k_2+b_3k_3)&&=x+h(b_1+b_2+b_3)f+h^2(b_2a_{21}+b_3a_{32})f'f\\ &&&\quad+h^3b_3a_{32}a_{21}f'^2f+\tfrac12h^3(b_2a_{21}^2+b_3a_{32}^2)f''f^2 \end{align} Now compare that with your formula for $x(t+h)$, leaving out all $t$ partial derivatives $$x(t+h)=x+hf+\tfrac12h^2f_xf+\tfrac16 h^3(f_{xx}f^2+f_x^2f)$$ to read off \begin{align} 1&=b_1+b_2+b_3\\ \tfrac12&=b_2a_{21}+b_3a_{32}\\ \tfrac13&=b_2a_{21}^2+b_3a_{32}^2\\ \tfrac16&=b_3a_{32}a_{21} \end{align} 4 equations for 5 parameters, so one degree of freedom left. Now insert $a_{21}=\tfrac12$, $a_{32}=\frac34$, $b_1=\frac29$, $b_2=\frac13$, $b_3=\frac49$.