TheoremProved

Taylor's theorem

Statement

If $f$ is $n+1$ times differentiable on an interval containing $a$ and $x$ , then $f(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k + R_n(x)$ , where the Lagrange remainder is $R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}$ for some $\xi$ strictly between $a$ and $x$ .

Why is it true?

A smooth function near a point is well approximated by a polynomial that matches its value and all its derivatives up to some order at that point; the remainder term measures exactly how good the approximation is and typically shrinks very fast as you add more terms.

Proof sketch

Fix $x$ and define $g(t)=f(x)-\sum_{k=0}^n\frac{f^{(k)}(t)}{k!}(x-t)^k$ and $h(t)=(x-t)^{n+1}$ . Then $g(x)=h(x)=0$ and, after telescoping, $g'(t)=-\frac{f^{(n+1)}(t)}{n!}(x-t)^n$ , while $h'(t)=-(n+1)(x-t)^n$ . Applying the Cauchy mean value theorem to $g,h$ on the interval between $a$ and $x$ gives $\frac{g(a)-g(x)}{h(a)-h(x)}=\frac{g'(\xi)}{h'(\xi)}$ for some $\xi$ between $a$ and $x$ , which simplifies exactly to $R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}$ .

Proved by

proved 1797Joseph-Louis Lagrange

Topics that use this theorem

Power series and Taylor series

Related theorems

Mean value theorem (Lagrange)Rolle's theorem

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

Brook Taylor (1715). Methodus Incrementorum Directa et Inversa
Joseph-Louis Lagrange (1797). Théorie des fonctions analytiques