Taylor polynomial

 

Taylor's Formula with the Integral (Cauchy) Form of the Remainder

Theorem

Assume $f$ has a continuous derivative of order $n+1$ in some interval containing $a$. Then, for every $x$ in this interval, we have the Taylor formula

$$f(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k+E_n(x),$$

where

$$E_n(x)=\frac{1}{n!}\int_{a}^{x} (x-t)^nf^{(n+1)}(t)dt.$$

Here, $E_n(x)$ may indicate the error involved in using the Taylor polynomial of order $n$ to approximate the given function.

 

For $n=1$

Assume $f$ has a continuous second derivative of order 2 in some neighborhood of $a$. Then, for every $x$ in this neighborhood, we have

$$f(x)=\sum_{k=0}^{1}\frac{f^{(k)}(a)}{k!}(x-a)^k+E_1(x)=f(a)+f'(a)(x-a)+E_1(x),$$

where

$$E_1(x)=\int_{a}^{x}(x-t)f''(t)dt.$$

We may write

$$ E_1(x) = f(x) - f(a) - f'(a)(x-a) = \int_{a}^{x} f'(t)dt - f'(a) \int_{a}^{x}dt = \int_{a}^{x} \left[ f'(t) - f'(a) \right] dt. $$

 

The last integral may be written as $\int_{a}^{x}udv$, where $u=f'(t)-f'(a),$ and $dv=dt.$ Now, $du/dt=f''(t)$ and $v=t-x,$ so the formula for integration by parts gives us

$$\begin{align*} E_1(x) &= \int_{a}^{x}udv = uv \rvert_{a}^{x} - \int_{a}^{x}vdu \\ &= (t-x)\left[ f'(t) - f'(a) \right]_{a}^{x} - \int_{a}^{x} (t-x)f''(t)dt \\ &= \int_{a}^{x}(x-t)f''(t)dt \end{align*}$$

This proves the theorem for $n=1.$

 

Proof by induction

Assume the theorem is true for some $n$ and prove it for $n+1$

Since

$$E_n(x)=f(x)-\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k,$$

we can write

$$\begin{align*} E_{n+1}(x) &= f(x)-\sum_{k=0}^{n+1}\frac{f^{(k)}(a)}{k!}(x-a)^k \\ &= f(x) - \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k - \frac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1} \\ &= E_n(x) - \frac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1}. \end{align*} $$

 

Now use the integral from for $E_n(x)$ and note that $\int_{a}^{x}(x-t)^ndt = -\frac{(x-t)^{n+1}}{(n+1)} |_{a}^{x} = \frac{(x-a)^{n+1}}{(n+1)}$

$$ \begin{align*} E_{n+1}(x) &= \frac{1}{n!} \int_{a}^{x}(x-t)^n f^{(n+1)}(t)dt - \frac{f^{(n+1)}(a)}{n!} \cdot \frac{(x-a)^{n+1}}{(n+1)} \\ &= \frac{1}{n!} \int_{a}^{x}(x-t)^n f^{(n+1)}(t)dt - \frac{f^{(n+1)}(a)}{n!} \int_{a}^{x}(x-t)^n dt \\ &= \frac{1}{n!} \int_{a}^{x} (x-t)^n \left[ f^{(n+1)}(t) - f^{(n+1)}(a) \right] dt \end{align*} $$

 

Again, the last integral may be written in the form $\int_{a}^{x}udv,$ where $u=f^{(n+1)}(t)-f^{(n+1)}(a)$ and $dv=(x-t)^n dt.$ Noting that $du=f^{(n+2)}(t)dt$ and $v=-\frac{(x-t)^{(n+1)}}{(n+1)}$, integration by parts gives us

$$ \begin{align*} E_{n+1}(x) &= \frac{1}{n!} \int_{a}^{x}udv = \frac{1}{n!} \left( \frac{(x-t)^{n+1}}{(n+1)} \left[ f^{(n+1)}(t) - f^{(n+1)}(a) \right]_a^x + \int_a^x \frac{(x-t)^{n+1}}{(n+1)}f^{(n+2)}(t)dt \right) \\ &= \frac{1}{n!}\int_a^x \frac{(x-t)^{n+1}}{(n+1)}f^{(n+2)}(t)dt = \frac{1}{(n+1)!} \int_a^x (x-t)^{n+1}f^{(n+2)}(t)dt \end{align*} $$

This complets the inductive step from $n$ to $n+1,$ so the theorem is true for all $n \geq 1.$

 

 

Theorem - Estimate of the error term

If the $(n+1)$st dervitive of $f$ satisfies the inequalities

$$m \leq f^{(n+1)}(t) \leq M \tag{1}$$

for all $t$ in some interval containing $a$, then for every $x$ in this interval we have the following estimates:

$$ m \frac{(x-a)^{n+1}}{(n+1)!} \leq E_n(x) \leq M \frac{(x-a)^{n+1}}{(n+1)!} \hspace{1cm} if \hspace{0.5cm} x>a,$$

and

$$ m \frac{(a-x)^{n+1}}{(n+1)!} \leq (-1)^{n+1}E_n(x) \leq M \frac{(a-x)^{n+1}}{(n+1)!} \hspace{1cm} if \hspace{0.5cm} x < a. $$ 

 

Proof. Assume that $x>a,$ then $E_n(x)$ is extended over $[a,x]$. The inequalities in (1) give us

$$ m\frac{(x-t)^n}{n!} \leq \frac{(x-t)^n}{n!}f^{(n+1)}(t) \leq M \frac{(x-t)^n}{n!} $$

$$ \hspace{5cm} \left( \text{Since } (x-t)^n \geq 0 \text{ for each } t \text{ in this interval.} \right) $$

Integrating from $a$ to $x$, we find that

$$ \frac{m}{n!} \int_a^x (x-t)^n dt \leq \frac{1}{n!} \int_a^x (x-t)^nf^{(n+1)}(t) dt = E_n(x) \leq \frac{M}{n!} \int_a^x (x-t)^n dt. \tag{2}$$

The substitution $u=x-t, du=-dt$ gives us

$$ \int_a^x (x-t)^ndt = \int_{x-a}^0 u^n (-du) \int_0^{x-a} u^n du = \frac{1}{n+1} u^{n+1} \vert_0^{x-a} = \frac{(x-a)^{n+1}}{n+1}. $$

Then, (2) reduces to 

$$ \frac{m}{n!} \frac{(x-a)^{n+1}}{n+1} \leq E_n(x) \leq \frac{M}{n!} \frac{(x-a)^{n+1}}{n+1} $$

$$ \Rightarrow \hspace{1cm} m \frac{(x-a)^{n+1}}{(n+1)!} \leq E_n(x) \leq M \frac{(x-a)^{n+1}}{(n+1)!}. $$

 

If $x<a$, the integration $E_n(x)$ takes place over the interval $[x,a]$. For each $t$ in this interval, we have $t \geq x,$ so $(t-x)^n=(-1)^n(x-t)^n= \geq 0.$ Hence, the inequalities in (1) give us

$$ m \frac{(-1)^n(x-t)^n}{n!} \leq \frac{(-1)^n(x-t)^n}{n!}f^{(n+1)}(t) \leq M \frac{(-1)^n(x-t)^n}{n!} $$

Integrating from $x$ to $a$, we obtain

$$ \frac{m(-1)^n}{n!} \int_x^a (x-t)^n dt \leq \frac{(-1)^n}{n!} \int_x^a (x-t)^n f^{(n+1)} dt \leq \frac{M(-1)^n}{n!} \int_x^a (x-t)^n dt. \tag{3}$$

Again, the substitution $u=x-t, du=-dt$ gives us

$$ \begin{align*} \int_x^a (x-t)^n dt &= \int_0^{x-a}u^n (-du) = \int_{x-a}^0 u^n du = \frac{1}{n+1}u^{n+1}\vert_{x-a}^0 \\ &= - \frac{(x-a)^{n+1}}{n+1} \end{align*} $$

Then,

$$ \frac{m(-1)^n}{n!} \int_x^a (x-t)^n dt = \frac{m(-1)^n}{n!}\frac{(-1)(x-a)^{n+1}}{n+1} = m \frac{(-1)^{n+1} (x-a)^{n+1}}{(n+1)!} = m\frac{(a-x)^{n+1}}{(n+1)!} $$

The middle part in (3) can be written in terms of $E_n(x)$ such that

$$\begin{align*} \frac{(-1)^n}{n!}\int_x^a (x-t)^n f^{(n+1)} dt &= \frac{(-1)^n}{n!} \left[ - \int_a^x (x-t)^nf^{(n+1)}dt \right] \\ &= (-1)^{n+1} \left[ \frac{1}{n!} \int_a^x (x-t)^n f^{(n+1)} dt \right] = (-1)^{n+1} E_n(x). \end{align*} $$

Therefore, (3) reduces to

$$ m \frac{(a-x)^{n+1}}{(n+1)!} \leq (-1)^{n+1}E_n(x) \leq M\frac{(a-x)^{n+1}}{(n+1)!}. $$