My textbook begins its proof of Taylor's theorem as follows:
$$f(x_0 + h) = f(x_0) + \int_{x_0}^{x_0 + h} f'(\tau) \ d \tau$$
Next, we write $d \tau = -d(x_0 + h – \tau)$ and integrate by parts$^1$ to give
$$f(x_0 + h) = f(x_0) + f'(x_0)h + \int_{x_0}^{x_0 + h} f''(\tau)(x_0 + h – \tau) \ d\tau$$
$^1$ Recall that integration by parts (the product rule for the derivative read backward) reads as
$$\int_a^b u \ dv = uv|^b_a – \int_a^b v \ du$$
Here we choose $u = f'(\tau)$ and $v = x_0 + h – \tau$.
If we compare
$$f(x_0 + h) = f(x_0) + \int_{x_0}^{x_0 + h} f'(\tau) \ d \tau$$
and
$$\int_a^b u \ dv = uv|^b_a – \int_a^b v \ du,$$
and we select $u = f'(\tau)$, then, notationally, we must have that $v = \tau$. But the author set $v = x_0 + h – \tau$, which makes no sense when we logically compare the mathematical statements?
It seems to me like the author has confused themselves and/or just misused notation?
I would greatly appreciate it if people could please clarify this.
Best Answer
I think what he is really doing here is changing variables. A more proper way to do this would be to let $s = -(x_0+h-\tau)$, then $ds = d\tau$. Then change the integral to an integral over $s$. Essentially this equates the first integral with an integral over different bounds (bounds on $s$).
Your statement "$d\tau=−d(x_0+h−\tau)$, which implies $\tau=\pm(x_0+h−\tau)$. So either $x_0=h=0$..." is not true. $d\tau=−d(x_0+h−\tau)$ implies that $\tau=-(x_0+h−\tau)+c$, where $c$ is an some constant, which we can set $c$ to $x_0+h$ to alleviate this issue. Notice that this is equivalent to defining $s$ as above and stating the fact that $\tau = s + x_0+h$, which is true from our construction.