Taylor s theorem

Topics in Calculus

Fundamental theorem | Function | Limits of functions | Continuity | Calculus with polynomials

Product rule | Quotient rule | Chain rule | Implicit differentiation | Taylor's theorem

Vector Calculus

Tensor Calculus

In calculus, Taylor's theorem, named after the mathematician Brook Taylor, who stated it in 1712, allows the approximation of a differentiable function near a point by a polynomial whose coefficients depend only on the derivatives of the function at that point. The precise statement is as follows: If n≥0 is an integer and f is a function which is n times continuously differentiable on the closed interval [a, x] and n+1 times differentiable on the open interval (a, x), then we have

$f(x) = f(a) + \frac{f'(a)}{1!}(x - a) + \frac{f^{(2)}(a)}{2!}(x - a)^2 + \cdots$

${} + \frac{f^{(n)}(a)}{n!}(x - a)^n + R$

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. Several expressions for R are available.

The exponential function y = e^x (red) and its corresponding Taylor's polynomial of degree 4 (blue)

The Lagrange form of the remainder term states that there exists a number ξ between a and x such that

$R = \frac{f^{(n+1)}(\xi)}{(n+1)!} (x-a)^{n+1}.$

This exposes Taylor's theorem as a generalization of the mean value theorem. In fact, the mean value theorem is used to prove Taylor's theorem with the Lagrange remainder term.

The Cauchy form of the remainder term is

$R = \int_a^x \frac{f^{(n+1)} (t)}{n!} (x - t)^n \, dt.$

This shows the theorem to be a generalization of the fundamental theorem of calculus.

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.

Proof

We first prove Taylor's theorem with the integral remainder term.

The fundamental theorem of calculus states that

$f(x) = f(a) + \int_a^x (x-t)^0 \, f'(t) \, dt.$

This proves the theorem for n = 0.

Integration by parts yields the case n = 1

$f(x) = f(a) +f'(a)\,(x-a)+\int_a^x (x-t)^1 \, f''(t) \, dt.$

By repeating this process, we may derive Taylor's theorem for higher values of n.

This can be formalized by applying the technique of induction. So, suppose that Taylor's theorem holds for a particular n, that is, suppose that

$f(x) = f(a) + \frac{f'(a)}{1!}(x - a) + \cdots + \frac{f^{(n)}(a)}{n!}(x - a)^n + \int_a^x \frac{f^{(n+1)} (t)}{n!} (x - t)^n \, dt. \qquad(*)$

We can again rewrite the integral using integration by parts. An antiderivative of (x − t)ⁿ as a function of t is given by −(x−t)ⁿ⁺¹ / (n + 1), so

$\int_a^x \frac{f^{(n+1)} (t)}{n!} (x - t)^n \, dt$

${} = - \left[ \frac{f^{(n+1)} (t)}{(n+1)n!} (x - t)^{n+1} \right]_a^x + \int_a^x \frac{f^{(n+2)} (t)}{(n+1)n!} (x - t)^{n+1} \, dt$

${} = \frac{f^{(n+1)} (t)}{(n+1)!} (x - a)^{n+1} + \int_a^x \frac{f^{(n+2)} (t)}{(n+1)!} (x - t)^{n+1} \, dt.$

Substituting this in (*) proves Taylor's theorem for n + 1, and hence for all nonnegative integers n.

The remainder term in the Lagrangian form can be derived by the mean value theorem in the following way:

$R = \int_a^x \frac{f^{(n+1)} (t)}{n!} (x - t)^n \, dt =f^{(n+1)}(\xi) \int_a^x \frac{(x - t)^n }{n!} \, dt$

The last integral can be solved immediately, which leads to

$R = \frac{f^{(n+1)}(\xi)}{(n+1)!} (x-a)^{n+1}$

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Categories: Calculus | Theorems

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