My textbook gives the following proof of the single-variable version of Taylor's theorem:
As promised, we begin with the Fundamental Theorem of Calculus, written in the form
$$f(x_0 + h) = f(x_0) + \int_{x_0}^{x_0 + h} f'(\tau) \ d \tau$$
Next, we write $d \tau = – d(x_0 + h – \tau)$ and integrate parts to give
$$f(x_0 + h) = f(x_0) = f'(x_0) h + \int_{x_0}^{x_0 + h} f''(\tau)(x_0 + h – \tau) \ d \tau,$$
which is the first-order Taylor formula. Integrating by parts again, we get
$$\int_{x_0}^{x_0 + h} f''(\tau)(x_0 + h – \tau) \ d\tau = – \dfrac{1}{2} \int_{x_0}^{x_0 + h} f''(\tau) d(x_0 + h – \tau)^2$$
$$= \dfrac{1}{2} f''(x_0) h^2 + \dfrac{1}{2} \int_{x_0}^{x_0 + h} f'''(\tau)(x_0 + h – \tau)^2 \ d \tau,$$
which, when substituted into the preceding formula, gives the second-order Taylor formula:
$$f(x_0 + h) = f(x_0) + f'(x_0) h + \dfrac{1}{2}f''(x_0) h^2 + \dfrac{1}{2} \int_{x_0}^{x_0 + h} f'''(\tau)(x_0 + h – \tau)^2 \ \ d\tau$$
I'm confused by the expression
$$- \dfrac{1}{2} \int_{x_0}^{x_0 + h} f''(\tau) d(x_0 + h – \tau)^2$$
Can someone explain what this means when the integral is without a $dx$, $dy$, $d\tau$, etc.? Is such a thing even a valid integral expression? And if there is no $d\tau$, then how are we supposed to know what we integrate the integrand with respect to?
I would greatly appreciate it if someone could please take the time to explain this.
Best Answer
I think you can safely ignore it, but what it means is: $$ \int_{x_0}^{x_0 + h} f''(\tau)(x_0 + h - \tau)d\tau = -\frac{1}{2}\int_{x_0}^{x_0 + h}f''(\tau)\frac{d}{d\tau}(x_0 + h - \tau)^2\; d\tau. $$ Given that (I hate this:) "the $d\tau$'s cancel", they have written the right-hand side above without either $d\tau$.