Taylor's Formula

Suppose now that $f(x)$ has $n$ derivatives in some interval about $x_0$. Suppose also that we have found a polynomial $P(x)$, of degree $(n-1)$, such that

f(x_0) = P(x_0),\quad f'(x_0) = P'(x_0),\quad \dots,\quad f^{(n-1)}(x_0) = P^{(n-1)}(x_0).

Then $f(x)-P(x)$ has a zero of multiplicity at least $n$ at $x_0$. By the approximation theorem, if we set $H(x)=(x-x_0{)}^n$ we have

$$f(x)-P(x)=\frac{H(x)}{n!}{f}^{(n)}(\xi ),$$

or

f(x) - P(x) = \frac{(x - x_0)^n}{n!} f^{(n)}(\xi), \tag{*}

where $\xi$ is between $x$ and $x_0$.

It remains to determine $P(x)$. To do so, we write

$$P(x)=a_0+a_1(x-x_0)+\cdots +a_{n-1}(x-x_0{)}^{n-1}$$

as a combination of powers of $(x-x_0)$, where the coefficients $a_i$ are yet to be determined. Now we wish to have $P(x_0)=f(x_0)$. But $P(x_0)=a_0$. Therefore $a_0=f(x_0)$. P'(x_0) = a_1, since the derivative of the constant term is $0$ and since a factor of $(x-x_0)$ remains in the derivatives of all terms except $a_1(x-x_0)$, which has derivative $a_1$. Therefore a_1 = f'(x_0). Likewise P''(x_0) = 2a_2 = f''(x_0), and in general $P^{(i)}(x_0)=i!a_1=f^{(i)}(x_0)$, or

$$a_i=\frac{f^{(i)}(x_0)}{i!}\cdot$$

Therefore

$$P(x)=\sum _{i=0}^{n-1}\frac{f^{(i)}(x_0)}{i!}(x-x_0{)}^i\cdot$$

Summarizing, we have

$$f(x)=\sum _{i=0}^{n-1}\frac{f^{(i)}(x_0)}{i!}(x-x_0{)}^i+\frac{(x-x_0{)}^n}{n!}{f}^{(n)}(\xi ),$$

where $\xi$ is between $x$ and $x_0$. This result is known as Taylor's Formula.

We can develop the formula by another approach which will supply some new information of interest. In this case we again assume that $f(x)$ has $n$ derivatives in an interval containing $a$ and . Then we let

F(x) = f(x) + \frac{b-x}{1!} f'(x) + \dots + \frac{(b-x)^{n-1}}{(n-1)!} f^{(n-1)}(x) \cdot

Then

\begin{aligned} F'(x) =& f'(x) - f'(x) + \frac{b-x}{1!} f''(x) - \frac{2(b-x)}{2!} f''(x) \\ &+ \frac{(b-x)^2}{2!} f'''(x)- \dots + \frac{(b-x)^{n-2}}{(n-2)!} f^{(n-1)}(x) \\ &- \frac{(n-1)(b-x)^{n-2}}{(n-1)!} f^{(n-1)}(x) + \frac{(b-x)^{n-1}}{(n-1)!} f^{(n)}(x)\\ &= \frac{(b-x)^{n-1}}{(n-1)!} f^{(n)}(x) , \end{aligned}

since the earlier terms all cancel in pairs. Now let

$$g(x)=f(b)-F(x)-C(b-x{)}^p,$$

where $1\le p\le n$ and $C$ is chosen so that $g(a)=0$. Since $F(b)=f(b)$, we also have $g(b)=0$. By Rolle's Theorem, there is a $\xi$, , such that

g'(\xi) = -F'(\xi) + pC(b-\xi)^{p-1} = 0.

Thus

C = \frac{F'(\xi)}{p(b-\xi)^{p-1}} = \frac{(b-\xi)^{n-p}}{p(n-1)!} f^{(n)}(\xi) ,

and

In particular, $g(a)=0$, so

$$f(b)-F(a)=\frac{(b-a{)}^p(b-\xi {)}^{n-p}}{p(n-1)!}{f}^{(n)}(\xi ),$$

or

f(b) = f(a) + \frac{(b-a)}{1!} f'(a) + \dots + \frac{(b-a)^{n-1}}{(n-1)!} f^{(n-1)}(a) + \frac{(b-a)^p(b-\xi)^{n-p}}{p(n-1)!} f^{(n)}(\xi) ,

for some $\xi$ between $a$ and $b$. If we replace $a$ by $x_0$ and $b$ by $x$, we see that the remainder we have here is exactly as before if $p=n$. This term is called the Lagrange remainder. For $p=1$, the remainder term is called the Cauchy remainder. In some cases it is convenient to use one, in some cases the other.

In general, the expression

f(x_0) + \frac{(x-x_0)}{1!} f'(x_0) + \frac{(x-x_0)^2}{2!} f''(x_0) + \dots

is called the Taylor's series for $𝒇\mathbf{(}𝒙\mathbf{)}$ about ${𝒙}_{\mathbf{0}}$. When the remainder after $n$ terms, that is, the difference between the first $n$ terms of the series and $f(x)$, approaches $0$ as $n$ becomes large, we say that the series converges to $f(x)$. As an example we shall show a series for $\arctan x$ about $x_0=0$ which converges to $\arctan x$ when $-1\le x\le 1$. This is really its Taylor's series about $0$. We shall use this series to demonstrate a very effective method for computing the decimal expansion of $\pi$.

Consider the series

$$\frac{x}{1}-\frac{x^3}{3}+\frac{x^5}{5}+\dots ,-1\leqq x\leqq 1.$$

Put

$$f_n(x)=\arctan x-(\frac{x}{1}-\frac{x^3}{3}+\cdots \pm \frac{x^{2n-1}}{2n-1}).$$

Then

\begin{aligned} f_n'(x) &= \frac{1}{1 + x^2} - \left( 1 - x^2 + x^4 - \dots \pm x^{2n-2} \right)\\ &= \frac{1}{1 + x^2} - \frac{1 - (-x^2)^n}{1 + x^2} , \end{aligned}

by the rule for the sum of a geometric series. This is simply

$$\frac{(-x^2{)}^n}{1+x^2}.$$

When $n$ is even, f_n'(x) \geqq 0. In this case $f_n(x)$ is monotone increasing. Since $f_n(0)=0$, $f_n(x)\geqq 0$ if $x\geqq 0$. Then, observing that when $n$ is even there are an even number of terms in our polynomial, we see that for $n$ even and $x\geqq 0$,

$$\arctan x\geqq \frac{x}{1}-\frac{x^3}{3}+\cdots -\frac{x^{2n-1}}{2n-1}.$$

For $n$ odd, f_n'(x) \leqq 0 and $f_n(x)\leqq 0$ for $x\geqq 0$, so

$$\arctan x\leqq \frac{x}{1}-\frac{x^3}{3}+\cdots +\frac{x^{2n-1}}{2n-1}.$$

Since $\arctan x$ lies between any two successive polynomials, the difference between a polynomial of $k$ terms and $\arctan x$ is at most the difference between this polynomial and the next one, of $(k+1)$ terms, or

$$\frac{x^{2k+1}}{2k+1}\leqq \frac{1}{2k+1},$$

since $0\leqq x\leqq 1$. Thus for $x$ in this range, the series converges to $\arctan x$. For $x\leqq 0$, we observe that $f_n(-x)=-f_n(x)$; since $f_n(-x)$ approaches zero, so does $f_n(x)$.

For $x=1$, this gives

$$\frac{\pi }{4}=\arctan 1=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dots .$$

Thus we have a way, but not a very economical one, of computing the decimal expansion of $\pi$. However, there is a closely related method which is very effective. Let $a=\arctan \frac{1}{5}$. Then

$$\tan 2a=\frac{2\tan a}{1-{\tan }^{2}a}=\frac{\frac{2}{5}}{1-\frac{1}{25}}=\frac{5}{12},$$

and once more,

$$\tan 4a=\frac{\frac{10}{12}}{1-\frac{25}{144}}=\frac{120}{119}.$$

Let $b=\arctan \frac{1}{239}$. Then

\begin{aligned} \tan (4a - b) &= \frac{\tan 4a - \tan b}{1 + \tan 4a \tan b} = \frac{\frac{120}{119} - \frac{1}{239}}{1 + \frac{120}{119} \cdot \frac{1}{239}}\\ &= \frac{120 \cdot 239 - 119}{119 \cdot 239 + 120} = \frac{119 \cdot 239 + 239 - 119}{119 \cdot 239 + 120} \\ &= 1 . \end{aligned}

Hence $4a-b=\frac{\pi }{4}$, $\pi =16a-4b$. Now

$$a=\frac{1}{5}-\frac{1}{3}{\left(\frac{1}{5}\right)}^3+\frac{1}{5}{\left(\frac{1}{5}\right)}^5-\dots ,$$

and

$$b=\frac{1}{239}-\frac{1}{3}{\left(\frac{1}{239}\right)}^3+\frac{1}{5}{\left(\frac{1}{239}\right)}^5-\dots .$$

Using this device, several hundred decimal places of $\pi$ have been computed.

Exercises