Taylor's theorem

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Topics in calculus

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Differentiation

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

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In calculus, Taylor's theorem, named after the mathematician Brook Taylor, who stated it in 1712, gives 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 most basic example is the approximation of the exponential function <math> \textrm{e}^x<math> near x = 0. Namely,

<math> \textrm{e}^x \approx 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^N}{N!}.<math>

The precise statement of the theorem 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

<math> 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_n <math>

Here, n! denotes the factorial of n, and Rn is a remainder term which depends on x and is small if x is close enough to a. Several expressions for Rn are available.

Missing image
Taylorspolynomialex.png
The exponential function y = ex (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

<math>
 R_n = \frac{f^{(n+1)}(\xi)}{(n+1)!} (x-a)^{n+1}.

<math>

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

<math>
 R_n(x) = \int_a^x \frac{f^{(n+1)} (t)}{n!} (x - t)^n \, dt.

<math>

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 Rn 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.

For complex functions analytic in a region containing a circle C surrounding a and its interior, we have a contour integral expression for the remainder

<math> R_n(x) = \frac{1}{2 \pi i}\int_C \frac{f(z)}{(z-a)^{n+1}(z-x)}dz<math>

valid inside of C.

Proof

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

The fundamental theorem of calculus states that

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

This proves the theorem for n = 0.

Integration by parts yields the case n = 1

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

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

<math>
 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(*) 

<math>

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

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

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

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

<math>
 R_n = \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.

<math>

The last integral can be solved immediately, which leads to

<math>
 R_n = \frac{f^{(n+1)}(\xi)}{(n+1)!} (x-a)^{n+1}.

<math>

External link

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