Integration and differentiation of an approximation to a function - order of approximation

Question

For my research I am working with approximations to functions which I then integrate or differentiate and I am wondering how this affects the order of approximation.

Consider as a minimal example the case of $e^x$ for which integration and differentiation doesn't change anything. Now if I would approximate this function with a second order taylor series I get: $$e^x\approx 1+x+\frac{x^2}{2}+O(x^3) \tag{1}$$

If I were to integrate this function I get: $$ \int 1+x+\frac{x^2}{2}+O(x^3) dx = C+x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^{?4?}) \tag{2}$$

I wrote $O^{?4?}$ because that is what my question is about: do I indeed get a higher order approximation when I do this integration or is it appropriate to cut-off the solution to the integral at $O(x^3)$, thus removing the $\frac{x^3}{6}$ term?

And what about differentiation? In that case I seem to lose an order of accuracy, is that indeed the case?

Answer 1

Suppose that $f(x)$ has a Taylor series expansion about $x=0$ with a radius of convergence $r>0$. For convenience we set $f(0)=1$.

We write $$ f(x)=1+xf^{(1)}(0)+\frac{x^{2}}{2}f^{(2)}(0)+\mathcal{O}% (x^{3})=1+xf^{(1)}(0)+\frac{x^{2}}{2}f^{(2)}(0)+g(x), $$ where, in a neighbourhood of $0$, $$ |x^{-3}g(x)|<c. $$ Then, for the primitive $F(x)$,

$$ F(x)-F(0)=\int_{0}^{x}dyf(y)=x\int_{0}^{1}duf(xu), $$ where $$ f(xu)=1+xuf^{(1)}(0)+\frac{(xu)^{2}}{2}f^{(2)}(0)+g(xu). $$ Then \begin{eqnarray*} \int_{0}^{1}duf(xu) &=&1+\frac{1}{2}xf^{(1)}(0)+\frac{x^{2}}{6}% f^{(2)}(0)+\int_{0}^{1}dug(xu) \\ &=&1+\frac{1}{2}xf^{(1)}(0)+\frac{x^{2}}{6}f^{(2)}(0)+\int_{0}^{1}du(xu)^{3}% \frac{g(xu)}{(xu)^{3}} \\ &=&1+\frac{1}{2}xf^{(1)}(0)+\frac{x^{2}}{6}f^{(2)}(0)+x^{3}% \int_{0}^{1}duu^{3}\frac{g(xu)}{(xu)^{3}} \\ \left\vert \int_{0}^{1}duu^{3}\frac{g(xu)}{(xu)^{3}}\right\vert &\leqslant &c\int_{0}^{1}duu^{3}=\frac{1}{4}c, \end{eqnarray*} so \begin{eqnarray*} F(x)-F(0) &=&x\{1+\frac{1}{2}xf^{(1)}(0)+\frac{x^{2}}{6}f^{(2)}(0)+x^{3}% \int_{0}^{1}duu^{3}\frac{g(xu)}{(xu)^{3}}\} \\ &=&x+\frac{1}{2}f^{(1)}(0)x^{2}+\frac{1}{6}f^{(2)}(0)x^{3}+x^{4}% \int_{0}^{1}duu^{3}\frac{g(xu)}{(xu)^{3}} \\ &=&x+\frac{1}{2}f^{(1)}(0)x^{2}+\frac{1}{6}f^{(2)}(0)x^{3}+\mathcal{O}% (x^{4}). \end{eqnarray*} For the derivative

\begin{eqnarray*} \partial _{x}f(x) &=&\partial _{x}\{1+xf^{(1)}(0)+\frac{x^{2}}{2}% f^{(2)}(0)+g(x)\} \\ &=&f^{(1)}(0)+xf^{(2)}(0)+\partial _{x}g(x). \end{eqnarray*} Now, according to l'Hôpital's rule,

$$ \lim_{x\rightarrow 0}\frac{g(x)}{x^{3}}=\lim_{x\rightarrow 0}\frac{\partial _{x}g(x)}{3x^{2}}, $$ so $$ \partial _{x}g(x)=\mathcal{O}(x^{2}) $$

Answer 2

This is indeed a very interesting question about the use of truncated series for integration and differentiation.

Taking you example of $e^x$, if we write $$f(x) =1+x+\frac{x^2}{2}+O\left(x^3\right)$$ from a formal point of view (at least, in my opinion), the derivative should write $$f'(x) =1+x+O\left(x^2\right)$$ and the antiderivative should write $$\int f(x) dx=C+x+\frac{x^2}{2}+\frac{x^3}{6}+O\left(x^4\right)$$