f(x)=f(0)+f'(0)*x+f''(x)/2！ *x^2+...+f(n)(0)/n！ * x n (McLaughlin formula, where n in the last term represents the n-order derivative)
Taylor's mean value theorem: if the function f(x) has a derivative up to n+ 1 in the open interval (a, b), when the function is in this interval, it can be expanded into the sum of (x-x.) polynomials and a remainder;
f(x)= f(x .)+f '(x .)(x-x .)+f ' '(x .)/2！ ? (x-x.)^2,+f'''(x.)/3！ ? (x-x.)^3+……+f(n)(x.)/n！ ? (x-x.)^n+Rn
Where rn = f (n+1) (ξ)/(n+1)! ? (x-X.) (n+ 1), where ξ is between x and x, and this remainder is called Lagrange remainder.
(Note: f(n) (x.) is the nth derivative of f (x.), not the product of f (n) and x. )
Prove: We know that f(x)=f(x.)+f'(x.)(x-x.)+α (the finite increment theorem derived from Lagrange's mean value theorem is lim δ x→ 0f (x.+δ x)-f (x.) = f' (x.) δ x), where the error is. Therefore, we need a polynomial that is accurate enough to estimate the error:
p(x)=a0+a 1(x-x.)+a2(x-x.)^2+……+an(x-x.)^n
Approximate the function f(x) and write the specific expression of its error f(x)-P(x). Let the function P(x) satisfy p (x.) = f (x x.), p p' (x.) = f'' (x x.), p' (x.) = f'' (x.), ..., p (n) (x.) = f (n). Obviously, P(x.)=A0, so A0 = F (X.); P'(x.)=A 1，a 1 = f '(x .)； P''(x.)=2！ A2，A2=f''(x.)/2！ ……P(n)(x.)=n！ An，An=f(n)(x.)/n！ . So far, the coefficients of many terms have been calculated, which are: p (x) = f (x.)+f' (x.) (x-x.)+f'' (x.)/2! ? (x-x.)^2+……+f(n)(x.)/n！ ? (x-x.)^n.
Next, the specific expression of the error is needed. Let Rn(x)=f(x)-P(x), so rn (x) = f (x)-p (x) = 0. So it can be concluded that rn (x.) = rn' (x.) = rn'' (x.) = ... = rn (n) (x.) = 0. According to Cauchy mean value theorem, we can get rn (x)/(x-x.) (n+1) = rn (x)-rn (x.)/(x-x.) (n+1)-0 = rn' (ξ 66) Continue to use Cauchy mean value theorem, Rn '(ξ 1)-Rn '(x .)/(n+ 1)(ξ 1-x .)n-0 = Rn ' '(9582)/n(n+)Rn(x)/(x-x .)(n+ 1)= Rn(n+ 1)(ξ)/(n+ 1) N+65438 is used continuously, where ξ is between X. and X. But Rn (n+1) (x) = f (n+1) (x)-p (n+1) (x), because P (n) (x) Ann, n! An is a constant, so P(n+ 1)(x)=0, so we get rn (n+1) (x) = f (n+1) (x). To sum up, the remainder rn (x) = f (n+1) (ξ)/(n+1)! ? (x-x.)^(n+ 1)。 Generally speaking, when expanding a function, it is for the need of calculation, so X often takes a fixed value, and then Rn(x) can also be written as Rn.
References:
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