Hello for the function
$$ y(x) = e^{\alpha x} \Leftrightarrow y' = \alpha y, y(0)=1 $$
I wanted to evaluate the error of the two following iterations
$$ \left\{ \begin{array}{l} y_k = y_{k-1} + \alpha y_{k-1} \delta t \\ y_0 = 1 \end{array} \right. \Rightarrow y_k = (1+\alpha \delta t)^k $$
and
$$ \left\{ \begin{array}{l} y_k = y_{k-1} + \alpha y_{k}\delta t \\ y_0 = 1 \end{array} \right. \Rightarrow (1 - \alpha \delta t)y_k = y_{k-1} \Rightarrow y_k = (1 - \alpha \delta t)^{-1} y_{k-1} \Rightarrow y_k = (1-\alpha \delta t)^{-k} $$
Now I'd like to evaluate the the errors using
$$ y(k\delta t) = e^{\alpha k \delta t} = 1 + \alpha k \delta t + \alpha^2 \frac{k^2}{2} \delta t^2 + ... $$
I can figure the taylor expansion of the first iteration (assuming $\delta t \rightarrow 0$), indeed I would get
$$ y_k = \sum_{l=0}^{k} \binom{k}{l} (\alpha \delta t)^l $$
In such a case $$ y(k\delta t) - y_k = O(\alpha \delta t) $$
In the second iteration though I was expecting less error (asymptotically at least, i.e. something like $O(\alpha^2 \delta t^2)$), Taylor expansion I managed to get, not sure if this is correct
$$ y_k = \sum_{l=0}^{+\infty} \frac{\Gamma(1-k)}{\Gamma(1-k-l)\Gamma(l+1)} (-1)^l(\alpha \delta t)^l = 1 + \alpha k \delta t + \frac{k (k+1)}{2}\alpha^2\delta t^2 + ... $$
however the entailed error in this case is the same, is there something I'm missing? or is what I've done correct?
You get easier results by comparing the powers in question in the exponents, that is as arguments of an exponential function after applying the appropriate logarithms. \begin{align} (1+αδt)^k=\exp\left(k\ln(1+αδt)\right) &=\exp\left(αkδt\left(1-\frac{αδt}2+\frac{(αδt)^2}3-\frac{(αδt)^3}4+...\right)\right) \\&=\exp\left(αt_k-\frac{α^2}2t_kδt+...\right), \\ (1-αδt)^{-k}=\exp\left(-k\ln(1-αδt)\right) &=\exp\left(αkδt\left(1+\frac{αδt}2+\frac{(αδt)^2}3+\frac{(αδt)^3}4+...\right)\right) \\&=\exp\left(αt_k+\frac{α^2}2t_kδt+...\right). \end{align} So you see, in both cases you get nearly the same first-order error, only with opposite signs.