Taylor's Theorem

Taylor's Theorem states that any infinitely differentiable function (including one where the derivative is 0) can be approximated by a series of polynomials.

Integral version
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} $$

The first equation is arrived at by letting $$u=f'(t)\,$$ and dv = dt; the second equation by noting that $$\int_a^x \, xf''(t) \,dt = xf'(x)-xf'(a)$$; the third just factors out some common terms.)

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}. $$

Mean value theorem
An alternative proof, which holds under milder technical assumptions on the function ƒ, can be supplied using the Cauchy mean value theorem. Let G be a real-valued function continuous on [a, x] and differentiable with non-vanishing derivative on (a, x). Let



F(t) = f(t) + \frac{f'(t)}{1!}(x-t) + \cdots + \frac{f^{(n)}(t)}{n!}(x-t)^n. $$

By Cauchy's mean value theorem,



\frac{F'(\xi)}{G'(\xi)} = \frac{F(x) - F(a)}{G(x) - G(a)} $$   (1)

for some &xi; &isin; (a, x). Note that the numerator F(x) − F(a) = Rn is the remainder of the Taylor polynomial for ƒ(x). On the other hand, computing F&prime;(t),



F'(t) = f'(t) - f'(t) + \frac{f(t)}{1!}(x-t) - \frac{f(t)}{1!}(x-t) + \cdots + \frac{f^{(n+1)}(t)}{n!}(x-t)^n = \frac{f^{(n+1)}(t)}{n!}(x-t)^n. $$

Putting these two facts together and rearranging the terms of (1) yields



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

which was to be shown.

Note that the Lagrange form of the remainder comes from taking G(t) = (x − t)n+1, and the given Cauchy form of the remainder comes from taking G(t) = (t − a).