I am required to derive Euler's method through Taylor's Theorem. I have been given the Taylor series for $y(t)$ as shown below. However I do not understand what point the Taylor series was derived. Isit at the point $a=t_i$ or $a=t_{i+1}?$
In this case, $y(t)$ is a unique solution to
$$ \frac{dy}{dt} = f(t, y),$$ $$ where \space\space a ≤ t ≤ b\space and\space\space y(a) = α.$$
The taylor series for this solution is
$$y(t_{i+1}) = y(t_{i}) + (t_{i+1} − t_{i})y′(t_i) + \frac{(t_{i+1} − t_i)^2}{2}y′′(ξ_i)$$
where i = 0, 1, 2, . . . , N − 1 and for some number $ξ_i$ in $(t_i, t_{i+1})$.
The (truncated) series is from the point $t_i$. Note that each of the polynomial terms is of the form $(t-t_i)$, where $t=t_{i+1}$. Additionally, the initial term is $y(t_i)$, so when $t=t_i$, the series evaluates to $y(t_i)$. If it were from $t_{i+1}$, then the initial term would be $y(t_{i+1})$, since that is what $y$ is at $t_{i+1}$.
You might think of it like equations of motion; $x(t)\approx x(t_0)+x'(t_0)(t-t_0)+x''(t_0)\frac{(t-t_0)^2}{2}$. You start, at $t=t_0$, from the position $x(t_0)$. As $t$ grows, $x(t)$ will initially move at its initial velocity, but the velocity will move in the direction of the initial acceleration. Regardless, when $t=t_0$, we are at the starting point, so the first term must be $x(t_0)$.