4.7 Sketch proof of Taylor’s theorem

(Not examinable).

The straight line segment from $(a,b)\,$ to $(a+h,b+k)\,$ is given by $x=a+th\,$ and $y=b+tk\,$ as the parameter ranges over $0\leq t\leq 1\,$. We introduce the function of a single variable

$$F(t)=f(a+th,b+tk)$$

so that $F(0)=f(a,b)\,$ and $F(1)=f(a+h,b+k)\,$ are the points that we wish to compare. The Maclaurin series for $F$, namely

$$F(t)=F(0)+{\frac{F^{\prime}(0)}{1!}}t+{\frac{F^{\prime\prime}(0)}{2!}}t^{2}+\cdots,$$

simplifies at $t=1\,$ to

$$F(1)=F(0)+{\frac{F^{\prime}(0)}{1!}}+{\frac{F^{\prime\prime}(0)}{2!}}+\cdots.$$

We now repeatedly apply the rule for calculating the derivative, to get: