Taylor's Theorem

Let $f$ is $n+1$ differentiable on $(a,b)$. Let $c,x \in (a,b)$. Then $\exists \zeta \in (c,x) \text{ s.t. }$:

$$f(x)= f(c) + \sum_{k=1}^{n}{\frac{f^{(k)}(c)}{k!}(x-c)^k} + \frac{f^{(n+1)}(\zeta)}{(n+1)!}{(x-c)}^{n+1}$$

Mean value theorem can be derived from taylor’s theorem when $n=0$. $$

Proof Hint

$$F(t)= f(t)+\sum_{k=1}^{n}{\frac{f^{(k)}(t)}{k!}(x-t)^k}$$$$G(t)=(x-t)^{n+1}$$

• Define $F,G$ as mentioned above
• Consider the interval $[c,x]$
• Use Cauchy’s mean value theorem for $F,G$ after making sure the conditions are met.

The above equation can be written like:

$$f(x)=T_n(x,c)+R_n(x,c)$$

Taylor Polynomial

This part of the above equation is called the Taylor polynomial. Denoted by $T_n(x,c)$. $$

$$T_n(x,c)= f(c) + \sum_{k=1}^{n}{\frac{f^{(k)}(c)}{k!}(x-c)^k}$$

Remainder

Derivative form

Denoted by $R_n(x,c)$. $$

$$R_n(x,c)= \frac{f^{(n+1)}(\zeta)}{(n+1)!}{(x-c)}^{n+1}$$

Integral form

$$R_n(x,c)= \frac{1}{n!} \int_c^x{f^{(n+1)}(t)(x-t)^n}\text{d}t$$

Proof Hint

• Method 1: Use integration by parts and mathematical induction.
• Method 2: Use Generalized IVT for Riemann Integrals where:
  • $F=f^{(n+1)}$
  • $G=(x-t)^n$