The Unapologetic Mathematician

Mathematics for the interested outsider

Taylor’s Theorem

I’ve decided I really do need one convergence result for Taylor series. In the form we’ll consider today, it’s an extension of the ideas in the Fundamental Theorem of Calculus.

Recall that if the function f has a continuous derivative f', then the Fundamental Theorem of Calculus states that

\displaystyle f(x)-f(x_0)=\int\limits_{x_0}^xf'(t)dt

Or, rearranging a bit

\displaystyle f(x)=f(x_0)+\int\limits_{x_0}^xf'(t)dt

That is, we start with the value at x_0, and we can integrate up the derivative to find how to adjust and find the value at the nearby point x. Now if f' is itself continuously differentiable we can integrate by parts to find

\displaystyle f(x)=f(x_0)+xf'(x)-x_0f'(x_0)-\int\limits_{x_0}^xtf''(t)dt

Then we use the FToC to replace f'(x)


And if f'' is itself continuously differentiable we can proceed to find

\displaystyle f(x)=f(x_0)+(x-x_0)f'(x_0)+\frac{1}{2}(x-x_0)^2f''(x_0)+\frac{1}{2}\int\limits_{x_0}^x(x-t)^2f'''(t)dt

Is this starting to look familiar?

At the nth step we’ve got

\displaystyle f(x)=\left(\sum\limits_{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k\right)+\int\limits_{x_0}^x\frac{f^{(n+1)}(t)}{n!}(x-t)^ndt

and if f^{(n+1)} is continuously differentiable we can integrate by parts and use the FToC to find

\displaystyle f(x)=\left(\sum\limits_{k=0}^{n+1}\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k\right)+\int\limits_{x_0}^x\frac{f^{(n+2)}(t)}{(n+1)!}(x-t)^{n+1}dt

The sum is the nth Taylor polynomial for f — the beginning of the Taylor series of f — at the point x_0, and the integral we call the “integral remainder term” R_n(x). For infinitely-differentiable functions we can define R_n for all n and get a sequence. The function f is then analytic if this sequence of errors converges to {0} in a neighborhood of x_0.

September 30, 2008 Posted by | Analysis, Calculus | 8 Comments