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Proof of Taylor formula Taylor Mean Value Theorem means that function f (x) is equal to the sum of polynomial PN (x) (that is, Taylor formula of order n of F (x)) and RN (x) (remainder of Taylor formula of order n of F (x)), and the remainder has the form [f (ξ) * (x-x0) ^ (n + 1)] / [(n + 1)!], so what needs to be proved is RN (x) = [f (ξ) * (x-x0) ^ (n + 1)] / [(n + 1)!].! why only need to prove RN?!
Many students can't understand the expression in the book
To prove the formula F (x) = PN (x) + [f (ξ) * (x-x0) ^ (n + 1)] / [(n + 1)!],
As long as we prove that f (x) - PN (x) = [f (ξ) * (x-x0) ^ (n + 1)] / [(n + 1)!],
Now we introduce the notation RN (x) = f (x) - PN (x)
So as long as we prove that RN (x) = [f (ξ) * (x-x0) ^ (n + 1)] / [(n + 1)!],
So as long as we prove that RN (x) / [(x-x0) ^ (n + 1)] = [f (ξ)] / [(n + 1)!],
After that, the Cauchy mean value theorem is used to prove the two functions on the left
How to prove Taylor formula
Taylor Mean Value Theorem: if the function f (x) has a derivative up to order n + 1 in the open interval (a, b), then when the function is in this interval, it can be expanded as the sum of a polynomial about (X-X.) and a remainder: F (x) = f (X.) + F '(X.) (X-X.) + F' '(X.) / 2! &; (X-X.) ^ 2, + F' '(X.) / 3! &; (X-X.)
Proof of Taylor formula?
In Taylor Mean Value Theorem, f (x) = f (X.) + F '(X.) (X-X.) + F' '(X.) / 2! (X-X.) ^ 2, + F' '' (X.) / 3! (X-X.) ^ 3 + +How to prove the equation f (n) (X.) / N! (X-X.) ^ n + RN? Why can f (x) be written like this?
Go to higher mathematics (I) or mathematical analysis (I) published by higher education press, where there is detailed proof
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