Taylor's theorem

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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. This result was first discovered 41 years earlier in 1671 by James Gregory.

1 Taylor's theorem in one variable
2 Taylor's theorem for several variables
3 Proof: Taylor's theorem in one variable
4 Proof: several variables
5 External links

[edit] Taylor's theorem in one variable

The exponential function y = e^x (red) and the corresponding Taylor's polynomial about a = 0 of degree four (blue)

The most basic example of Taylor's theorem is the approximation of the exponential function $e x$ near x = 0. Namely,

$\textrm{e}^x \approx 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!}.$

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

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

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

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

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

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

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

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_n approaches zero as n approaches ∞; those functions can be expressed as a Taylor series in a neighbourhood 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

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

valid inside of C.

[edit] Taylor's theorem for several variables

Taylor's theorem can be generalized to several variables as follows. Let B be a ball in R^N centered at a point a, and f be a real-valued function defined on the closure $\bar{B}$ having n+1 continuous partial derivatives at every point. Taylor's theorem asserts that for any $x\in B$ ,

$f(x)=\sum_{|\alpha|=0}^n\frac{D^\alpha f(a)}{\alpha!}(x-a)^\alpha+\sum_{|\alpha|=n+1}R_{\alpha}(x)(x-a)^\alpha$

where the summation extends over multi-indices α (this formula uses the multi-index notation).

The remainder terms satisfy the inequality

$|R_{\alpha}(x)|\le\sup_{y\in\bar{B} }\left|\frac{D^\alpha f(y)}{\alpha!}\right|$

for all α with |α|=n+1. As was the case with one variable, the remainder terms can be described explicitly. See the proof for details.

[edit] Proof: Taylor's theorem in one variable

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

The fundamental theorem of calculus states that

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

which can be rearranged to:

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

Now we can see that an application of Integration by parts yields:

$\begin{align} f(x) &= f(a)+xf'(x)-af'(a)-\int_a^x \, tf''(t) \, dt \\ &= f(a)+\int_a^x \, xf''(t) \,dt+xf'(a)-af'(a)-\int_a^x \, tf''(t) \, dt \\ &= f(a)+(x-a)f'(a)+\int_a^x \, (x-t)f''(t) \, dt. \end{align}$

Another application yields:

$f(x)=f(a)+(x-a)f'(a)+ \frac 1 2 (x-a)^2f''(a) + \frac 1 2 \int_a^x \, (x-t)^2f'''(t) \, dt.$

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 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

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

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

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

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

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

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

The last integral can be solved immediately, which leads to

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

[edit] Proof: several variables

Let x=(x₁,...,x_N) lie in the ball B with center a. Parametrize the line segment between a and x by u(t)=a+t(x-a). We apply the one-variable version of Taylor's theorem to the function f(u(t)):

$f(x)=f(u(1))=f(a)+\sum_{k=1}^n\left.\frac{1}{k!}\frac{d^k}{dt^k}\right|_{t=0}f(u(t))\ +\ \int_0^1 \frac{(1-t)^n }{n!} \frac{d^{n+1}}{dt^{n+1}} f(u(t)) dt.$

By the chain rule for several variables,

$\frac{d^k}{dt^k}f(u(t)) = \frac{d^k}{dt^k}f(a+t(x-a))=\sum_{|\alpha|=k}\left(\begin{matrix}k\\ \alpha\end{matrix}\right)(D^\alpha f)(a+t(x-a))\cdot (x-a)^\alpha$

where $\left(\begin{matrix}k\\ \alpha\end{matrix}\right)$ is the multinomial coefficient for the multi-index α. Since $\frac{1}{k!}\left(\begin{matrix}k\\ \alpha\end{matrix}\right)=\frac{1}{\alpha!}$ , we get

$f(x)= f(a)+\sum_{|\alpha|=1}^n\frac{1}{\alpha!} (D^\alpha f) (a)(x-a)^\alpha+\sum_{|\alpha|=n+1}\frac{n+1}{\alpha!} (x-a)^\alpha \int_0^1 (1-t)^n (D^\alpha f)(a+t(x-a))dt.$