Say I have two functions $f(x)$ and $g(x)$, both of which I will be approximating with Taylor series $T_f(x)$ and $T_g(x)$ respectively. Lets say $f(x)$ is order $O(x^{n_1})$ and $T_f(x)$ has error of order $O(x^{m_1})$. Likewise $g(x)$ is order $O(x^{n_2})$ and $T_g(x)$ has error of order $O(x^{m_2})$.
Given that, what is the order of the error for $T_f(x)T_g(x)$?
What I have thought of so far is that $f(x)g(x) = (T_f(x)+E_f(x))(T_g(x)+E_g(x))= T_f(x)T_g(x) + E_f(x)T_g(x) + E_g(x)T_f(x) + E_f(x)E_g(x)$
and if the orders multiply then $E_{total}=\max\{x^{n_2+m_1},x^{n_1+m_2},x^{m_2+m_1}\}$
But that looks like the error would go down, not up like it should be.
When you multiply two truncated series you get an approximation of the product function which is as good as the lowest one: $$ \eqalign{ & \left( {a_{\,0} + a_{\,1} x + a_{\,2} x^{\,2} + \ldots + O\left( {x^{\,m} } \right)} \right)\left( {b_{\,0} + b_{\,1} x + b_{\,2} x^{\,2} + \ldots + O\left( {x^{\,n} } \right)} \right) = \cr & = c_{\,0} + c_{\,1} x + c_{\,2} x^{\,2} + \ldots + b_{\,0} O\left( {x^{\,m} } \right) + \ldots + a_{\,0} O\left( {x^{\,n} } \right) = \cr & = c_{\,0} + c_{\,1} x + c_{\,2} x^{\,2} + \ldots + O\left( {x^{\,\min \,(m,\,n)} } \right) \cr} $$