I know that Riemann integration is taught after Taylor's Theorem in most Real Analysis courses, but it is an interesting result that one may in fact use integration by parts on the remainder term to come up with a similar result, though with a larger coefficient in the remainder term. The proof is a 2 sentence inductive argument! Quite a nice tradeoff to teach students a simple proof for what is such an important result.
@brightsideofmaths11 ай бұрын
Yes, I like that as well :)
@andrewxiwu11 ай бұрын
Great video! A perhaps more intuitive proof (that doesn't feel "somewhat tricky"), we can consider these two functions (for fixed x_0): D(t) = f(x_0 + t) - T_n(t) # T_n(t) is the Taylor until degree n G(t) = t^{n+1} Then we are interested in studying the difference term D(h), and we can apply the generalized/extended MVT to (D(h) - D(0))/(G(h) - G(0)). This will give us a form where we can then "apply induction on n".