# Taylor series

The Taylor expansion theorem, in a sense, asks how well functions can be approximated by polynomials, that is, for a given function~$f$, can we find coefficients~$c_i$ with $i=1,\ldots,n$ so that $f(x)\approx c_0+c_1x+c_2x^2+\cdots+c_nx^n.$ This question obviously needs to be refined. What do we mean by approximately equal'? This approximation formula can not hold for all functions~$f$ and all~$x$: the function $\sin x$ is bounded for all~$x$, but any polynomial is unbounded for $x\rightarrow \pm\infty$, so any polynomial approximation to the $\sin x$ function is unbounded. Clearly we can only approximate on an interval.
We will show that a function~$f$ with sufficiently many derivatives can be approximated as follows: if the $n$-th derivative~$f^{(n)}$ is continuous on an interval~$I$, then there are coefficients $c_0,\ldots,c_{n-1}$ such that $\forall_{x\in I}\colon |f(x)-\sum_{i Now we need to be a bit more precise. Cauchy's form of Taylor's theorem says that \[ f(x) = f(a)+\frac1{1!}f'(a)(x-a)+\cdots+\frac1{n!}f^{(n)}(a)(x-a)^n +R_n(x)$ where the `rest term' $R_n$ is $R_n(x) = \frac1{(n+1)!}f^{(n+1)}(\xi)(x-a)^{n+1} \quad\hbox{where \xi\in(a,x) or \xi\in(x,a) depending.}$ If $f^{(n+1)}$ is bounded, and $x=a+h$, then the form in which we often use Taylor's theorem is $f(x) = \sum_{k=0}^n \frac1{k!}f^{(k)}(a)h^k+O(h^{n+1}).$ We have now approximated the function~$f$ on a certain interval by a polynomial, with an error that decreases geometrically with the inverse of the degree of the polynomial.
For a proof of Taylor's theorem we use integration by parts. First we write $\int_a^xf'(t)dt=f(x)-f(a)$ as $f(x) = f(a)+\int_a^xf'(t)dt$ Integration by parts then gives $\begin{array}{r@{=}l} f(x) &f(a)+[xf'(x)-af'(a)]-\int_a^xtf''(t)dt\\ &f(a)+[xf'(x)-xf'(a)+xf'(a)-af'(a)]-\int_a^xtf''(t)dt\\ &f(a)+x\int_a^xf''(t)dt+(x-a)f'(a)-\int_a^xtf''(t)dt\\ &f(a)+(x-a)f'(a)+\int_a^x(x-t)f''(t)dt\\ \end{array}$ Another application of integration by parts gives $f(x)=f(a)+(x-a)f'(a)+\frac12(x-a)^2f''(a) +\frac12 \int_a^x(x-t)^2f'''(t)dt$ Inductively, this gives us Taylor's theorem with $R_{n+1}(x) = \frac1{n!}\int_a^x(x-t)^nf^{(n+1)}(t)dt$ By the mean value theorem this is $\begin{array}{r@{=}l} R_{n+1}(x) &\frac1{(n+1)!}f^{(n+1)}(\xi)\int_a^x(x-t)^nf^{(n+1)}(t)dt\\ &\frac1{(n+1)!}f^{(n+1)}(\xi)(x-a)^{n+1} \end{array}$