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I understand that the Taylor series isn't always equal to f(x) for each x, so we put Rn at the end as the remainder term (note that a + h = x).

f(a+h) = f(a) + [tex]\frac{h}{1!}[/tex]*f'(a) + [tex]\frac{h^2}{2!}[/tex]*f''(a)+⋯+[tex]\frac{h^n}{n!}[/tex] f[tex]^{(n)}[/tex] (a) +Rn

So Rn is f(x) minus it's Taylor series.

But then we try to approximate how big Rn actually is. For fixed values of a and x, we look at a new function that looks like this (t is any number, p [tex]\in[/tex] (1, 2, 3, ..., n)):

F(t) = f(x) - f(t) - [tex]\frac{x - t}{1!}[/tex]*f'(t) - [tex]\frac{(x - t)^{2}}{2!}[/tex]*f''(t) - ... - [tex]\frac{(x - t)^{^n}}{n!}[/tex]*f[tex]^{(n)}[/tex](t) - Rn(x)*([tex]\frac{(x - t)^{p}}{(x - a)^{p}}[/tex]^{p}

From here on we say that F(a) = F(x) = 0 and see that, following Rolle, there should be a value w between a and x for which F'(w) = 0 and that's how we end up with a formula for Rn.

What bothers me here: What is F(t)? And if I compare it to the first formula, where did that fraction of ((x-t)/(x-a))^p next to Rn come from?

Thank you for your time!