Taylor's Theorem

Theorem (Taylor’s Theorem)

Suppose

• $f$ is a real function on $[a,b]$
• $n$ is a positive integer
• $f^{(n-1)}$ is continuous on $[a,b]$
• $f^{(n)}(t)$ exists for every $t\in (a,b)$

Let $\alpha ,\beta$ be distinct points of $[a,b]$, and define

$$P(t)=\sum _{k=0}^{n-1}\frac{f^{(k)}(\alpha )}{k!}(t-\alpha {)}^k$$

Then there exists a point $x$ between $\alpha$ and $\beta$ such that

$$f(\beta )=P(\beta )+\frac{f^{(n)}(x)}{n!}(\beta -\alpha {)}^n$$

For $n=1$, this is just the Theorem (Lagrange Mean Value Theorem). In general, the theorem shows that $f$ can be approximated by a polynomial of degree $n-1$, and that the second equation allows us to estimate the error if we know bounds on $|f^{(n)}(x)|$.

Proof: Let $M$ be defined by

$$f(\beta )=P(\beta )+M(\beta -\alpha {)}^n$$

and put

$$g(t)=f(t)-P(t)-M(t-\alpha {)}^n(a\le t\le b)$$

We have to show that $M=f^{(n)}(x)/n!$ for some $x$ between $\alpha$ and $\beta$.

Recall that Taylor’s Theorem is a sum. So we need to find that the next element is possible. Note that $g$ is defined this way so that $g(\alpha )=g(\beta )=0$.

• $g(\beta )=0$ since rearranging the definition of $M$ is the same as $g(\beta )$ and it is equal to $0$.
• $g(\alpha )=0$ since the second term is $P(\alpha )=f(\alpha )$. $$P(\alpha )=\sum _{k=0}^{n=1}\frac{f^{(k)}(\alpha )}{k!}(\alpha -\alpha {)}^k=\frac{f^{(0)}(\alpha )}{0!}(\alpha -\alpha {)}^0+\sum _{k=1}^{n-1}{0}^k=f(\alpha )$$ and $g(\alpha )=f(\alpha )-f(\alpha )-0=0$.

By the original definition and the equation above,

$$g^{(n)}(t)=f^{(n)}(t)-n!M(a<t<b)$$

Hence the proof will be complete if we can show that $g^{(n)}(x)=0$ for some $x$ between $\alpha ,\beta$. Since $P^{(k)}(\alpha )=f^{(k)}(\alpha )$ for $k=0,\dots ,n-1$, we have

$$g(\alpha )=g^{\prime }(\alpha )=\cdots =g^{(n-1)}(\alpha )=0$$

After $n$ steps we conclude that $g^{(n)}(x_n)=0$ for some $x_n$ between $\alpha$ and $x_{n-1}$, that is, between $\alpha$ and $\beta$.