Redirected from Taylors theorem
f(x) = f(a) + \frac{f'(a)}{1!}(x  a) + \frac{f^{(2)}(a)}{2!}(x  a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x  a)^n + R</math>
Here, n! denotes the factorial of n, and R is a remainder term which depends on x and is small if x is close enough to a. Three expressions for R are available. Two are shown below:
R = \frac{f^{(n+1)}(\xi)}{(n+1)!} (xa)^{n+1}</math>
where ξ is a number between a and x, and
R = \int_a^x \frac{f^{(n+1)} (t)}{n!} (x  t)^n \, dt</math>
If R is expressed in the first form, the socalled Lagrange form, Taylor's theorem is exposed as a generalization of the mean value theorem (which is also used to prove this version), while the second expression for R shows the theorem to be a generalization of the fundamental theorem of calculus (which is used in the proof of that version).
For some functions f(x), one can show that the remainder term R approaches zero as n approaches ∞; those functions can be expressed as a Taylor series in a neighborhood of the point a and are called analytic.
Taylor's theorem (with the integral formulation of the remainder term) is also valid if the function f has complex values or vector values. Furthermore, there is a version of Taylor's theorem for functions in several variables.
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