calculus - Why Does The Taylor Remainder Formula Work?

I've been studying calculus on my own and have come across Taylor series. It is very intuitive until I came across the remainder part of the formula where things got fuzzy. I understand why the remainder exists but not the mathematical description. Why is the value being plugged into the derivative of the remainder some number between $x$ and $a$? What is the connection to the mean value theorem? Is the remainder used to bound the function? Lastly and my most important question is what is the intuitive (for me most likely a geometrical approach would be great) for how this remainder is derived. I've spent a long time trying to figure it out for myself and looking online but it seems I'm missing something for there are virtually no questions being asked about this.

Thanks for you time,
Jackson

Answer

Perhaps not quite the way you are looking for, but:

You can derive Taylor's theorem with the integral form of the remainder by repeated integration by parts:
$$ f(x)-f(a) = \int_a^x f'(t) \, dt = \left[-(x-t)f'(t) \right]_a^x + \int_a^x (x-t) f''(t) \, dt \\
= (x-a)f'(a) + \int_a^x (x-t) f''(t) \, dt, $$

and so on, integrating the $(x-t)$ and differentiating the $f$ each time, to arrive at
$$ R_N = \int_a^x \frac{(x-t)^N}{N!} f^{(N+1)}(t) \, dt. \tag{1} $$
Interpretation for this is simply that integrating by parts in the other direction will give you back precisely $f(x) - f(a) - \dotsb - \frac{1}{N!}(x-a)^N f^{(N)}(a)$.

Now, we can get from (1) to the Lagrange and Cauchy forms of the remainder by using the Mean Value Theorem for Integrals, in the form:

Let $g,h$ be continuous, and $g>0$ on $(a,b)$. Then $\exists c \in (a,b)$ such that
$$ \int_a^b h(t) g(t) \, dt = h(c) \int_a^b g(t) \, dt. $$

(this is easy if you think about weighted averages and the usual Mean Value Theorem).

Applying this to (1) with $h=f$, $g(t)=(x-t)^N/N!$ gives
$$ R_N = f^{(N+1)}(c)\frac{(x-a)^{N+1}}{(N+1)!}, $$
which is the Lagrange form of the remainder; using $h(t)=f(t)(x-t)^N/N!$, $g(t)=1$ gives
$$ R_N = f^{(N+1)}(c')\frac{(x-c')^N}{N!}(x-a), $$
which is the Cauchy form of the remainder.

The weighted averages mentioned above are a way to think about what we did here: we take an average of $\frac{(x-t)^N}{N!} f^{(N+1)}(t)$ over $[a,x]$, and use the MVT to equate this to a value of the function at a specific point: how much of the function we count as in the weighting affects what answer we obtain.