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Taylor's Theorem

Is it possible to express transcendental function $f(x)$ such as $e^x$, $\log{x}$, $\sin{x}$, $\tan^{-1}{x}$ using polynomials? The next theorem answers such a question.

Taylor's Theorem

Theorem 2..14   If $f(x)$ is class $C^{n}$ on $[a,b]$. Then there exists $\xi$ in $(a,b)$ such that

$\displaystyle f(b) = \underbrace{f(a) + f^{\prime}(a)(b-a) + \cdots + \frac{f^{... ...1)!}(b-a)^{n-1}}_{\rm Taylor polynomial} + \underbrace{R_{n}}_{\rm Remainder}, $

$\displaystyle R_{n} = \underbrace{\frac{f^{(n)}(\xi)}{n!}(b-a)^{n}}_{\rm Lagrange's Remainder}. $

NOTE Note that for $n=1$, we have $f(b) = f(a) + R_1$, where $R_1 = f'(\xi)(b-a)$. Thus for $n=1$, Taylor's theorem is the same as Mean Value Theorem. For $n = 2$, we have $f(b) = f(a) + f'(a)(b-a) + R_2$, where $R_2$ is the difference of the value of line $y = f'(a)(x-a) + f(a)$ and function $f(x)$ at $x = b$. In other words, $R_2$ is an error caused by approximation of the value of $f$ by the line. Similarly, $R_3$ is an approximation error of $f$ by a quadratic polynomial.

Proof Let $K$ be an expression satisfying

$\displaystyle f(b) = f(a) + f^{\prime}(a)(b-a) + \cdots + \frac{f^{(n-1)}(a)}{(n-1)!}(b-a)^{n-1} + \frac{(b-a)^{n}}{n!}K $

Let

$\displaystyle F(x) = f(b) - f(x) - \sum_{k=1}^{n-1}\frac{f^{(k)}(x)}{k!}(b-x)^{k} - \frac{(b-x)^{n}}{n!}K $

Then $F(a) = 0, F(b) = 0$. Thus $F(x)$ satisfies the condition of Rolle's Theorem . Thus
$\displaystyle F^{\prime}(\xi)$ $\displaystyle =$ $\displaystyle -f^{\prime}(\xi) - \sum_{k=2}^{n}\frac{f^{(k)}(\xi)}{(k-1)!}(b-\xi)^{k-1} + \sum_{k=1}^{n-1}\frac{f^{(k)}(\xi)}{(k-1)!}(b-\xi)^{k-1}$  
  $\displaystyle +$ $\displaystyle \frac{(b-\xi)^{n-1}}{(n-1)!}K$  
  $\displaystyle =$ $\displaystyle -f^{\prime}(\xi) + f^{\prime}(\xi) - \frac{f^{n}(\xi)}{(n-1)!}(b-\xi)^{n-1} + \frac{(b-\xi)^{n-1}}{(n-1)!}K = 0$  

Therefore,

$\displaystyle K = f^{(n)}(\xi) \ensuremath{\ \blacksquare} $

In Taylor's theorem with $a = 0$ is called Maclaurin's Theorem. Set $b = x$. Then we have

$\displaystyle f(x) = f(0) + f^{\prime}(0)x + \cdots + \frac{f^{(n-1)}(0)}{(n-1)!}x^{n-1} + R_{n} = \sum_{k=0}^{n-1}\frac{f^{(k)}(0)}{k!}x^k + R_n,$

where$\displaystyle ,\ R_{n} = \frac{f^{(n)}(\xi)}{n!}x^{n} = \frac{f^{(n)}(\theta x)}{n!}x^{n}, \ 0 < \theta < 1. $

Now error estimate is given by $\vert R_n\vert$, where

$\displaystyle \vert R_{n}\vert = \vert\frac{f^{(n)}(\theta x)}{n!}x^{n}\vert \l... ...ax_{\theta \in [0,1]}\vert f^{(n)}(\theta x)\vert)\frac{\vert x\vert^{n}}{n!}. $

Example 2..15   Find a Taylor polynomial and error estimate of the following function expanded around $a = 0$.

$\displaystyle f(x) = e^{x}$

SOLUTION Since $f^{(n)}(x) = e^{x}$, we have $f^{(n)}(0) = 1$. Find a Taylor polynomial around $a = 0$, we have

$\displaystyle P(x) = \sum_{k=0}^{n-1}\frac{1}{k!}x^k = 1 + x+ \frac{x^{2}}{2!} + \cdots + \frac{x^{n-1}}{(n-1)!} . $

We next find error estimate. Since

$\displaystyle R_{n} = \frac{f^{(n)}(\theta x)}{n!}x^n = \frac{e^{\theta x} x^{n}}{n!} $

we have

$\displaystyle \vert R_{n}\vert \leq (\max_{\theta \in (0,1)}\vert e^{\theta x}\... ...n!} \leq \vert e^{x}\vert\frac{\vert x\vert^{n}}{n!}\ensuremath{\ \blacksquare}$

Exercise 2..15   Find a Taylor polynomial and error estimate of the following function expanded around $a = 0$

$\displaystyle f(x) = \cos{x}$

SOLUTION Since $f^{(n)}(x) = \cos(x+\frac{n\pi}{2})$, we have $f^{(n)}(0) = \cos(\frac{n\pi}{2})$. Thus Taylor polynomial around $a = 0$ is $\displaystyle{P(x) = \sum_{k=0}^{n-1}\frac{f^{(k)}(0)}{k!}x^k.}$ Now we divide this into two cases.
case 1. $k$ is even. Let $k = 2m$. Then $f^{(2m)}(0) = \cos(\frac{2m\pi}{2}) = \cos(m\pi) = (-1)^{m}$
case 2. $k$ is odd. Let $k = 2m+1$. Then $f^{(2m+1)}(0) = \cos(\frac{(2m+1)\pi}{2}) = 0$. Thus
$\displaystyle P(x)$ $\displaystyle =$ $\displaystyle \sum_{k=0}^{n-1}\frac{f^{(k)}(0)}{k!}x^k = \sum_{m=0}^{\frac{n-1}{2}}\frac{(-1)^m}{(2m)!}x^{2m}.$  

We next find error estimate.

$\displaystyle R_{n} = \frac{\cos({\theta x} + \frac{n\pi}{2}) x^{n}}{n!}, \ 0 < \theta < 1$

Thus error estimate is

$\displaystyle \vert R_{n}\vert \leq (\max_{\theta \in (0,1)}\vert\cos(\theta x ... ...ert x\vert^{n}}{n!} \leq \frac{\vert x\vert^{n}}{n!}\ensuremath{\ \blacksquare}$

MacLaurin Series Expansion
Suppose that $f(x)$ is infinitely many times differentiable function on an interval containing $x = 0$. The by MacLaurin's theorem, we have

$\displaystyle f(x) = \sum_{k=0}^{n-1} \frac{f^{(k)}(0)}{(k)!}x^{k} + R_{n} $

If $R_{n} \rightarrow 0 \ (n \rightarrow \infty)$, then we can express $f(x)$ as

$\displaystyle f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{(n)!}x^{n}$

In this case, the right-hand side is called MacLaurin Series Expansion of $f(x)$.
MacLaurin Series Expansion of Basic Functions

Theorem 2..15  

1. $\displaystyle{e^{x} = \sum_{n=0}^{\infty}\frac{1}{n!}x^n}, \ (-\infty < x < \infty)$
2. $\displaystyle{\sin{x} = \sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n + 1)!}}, \ (-\infty < x < \infty)$
3. $\displaystyle{\cos{x} = \sum_{n=0}^{\infty} (-1)^{n}\frac{x^{2n}}{(2n)!} },\ (-\infty < x < \infty)$
4. $\displaystyle{\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n}, \ (-1 < x < 1)$
5. $\displaystyle{\log(1+x) = \sum_{n=0}^{\infty} (-1)^{n-1}\frac{x^{n}}{n}}, \ (-1 < x \leq 1))$
6. $\displaystyle{(1+x)^{\alpha} = \sum_{n=0}^{\infty}\frac{\alpha(\alpha-1)\cdots(\alpha -n+1)}{n!}x^{n}, (-1 < x < 1)}$

NOTE To show MacLaurin series expansion, we need to show $\lim_{n \to \infty}R_n = 0$. But it is not easy to show $\lim_{n \to \infty}R_n = 0$ by using Lagrange's Remainder $R_n$. So, we use different method. Suppose we express a MacLaurin series expansion of $f(x)$ as $\sum_{n=0}^{\infty}b_n$. Then showing $\vert R_n\vert$ approachs 0 is the same as showing $\sum_{n=0}^{\infty}b_n$ converges. To show $\sum_{n=0}^{\infty}b_n$ converges, it is useful Limit Ration Test.

Limit Ration Test

Theorem 2..16   Suppose $\sum_{n=0}^{\infty}b_n$ be a nonnegative series. If $\lim_{n \to \infty}\vert\frac{b_{n+1}}{b_n}\vert < 1$, then $\sum_{n=0}^{\infty}b_n$ converges.

Example 2..16   Find a MacLaurin series expansion of the following functions.
1. $f(x) = e^{x}$ 2. $f(x) = \sin{x}$

SOLUTION 1. Let $f(x) = e^x$. Then since $f^{(n)}(x) = e^x$, we have $f^{(n)}(0) = 1$. Thus Taylor polynomial $P(x)$ for $a = 0$ is

$\displaystyle P(x) = \sum_{k=0}^{n-1}\frac{f^{(k)}(0)}{k!}x^k = \sum_{k=0}^{n-1}\frac{1}{k!}x^k.$

Let $b_n = \frac{f^{(n)}(0)}{n!}x^n = \frac{1}{n!}x^n$. Then apply the limit ration test.

$\displaystyle \lim_{n \to \infty}\vert\frac{b_{n+1}}{b_n}\vert = \lim_{n \to \i... ...}\cdot \frac{n!}{x^n}\vert = \lim_{n \to \infty}\vert\frac{x}{n+1}\vert = 0 < 1$

Thus for all $x$, $\sum_{k=0}^{\infty}\frac{1}{k!}x^k$ converges. Therefore $\lim_{n \to \infty}R_n = 0$ and,

$\displaystyle e^{x} = \sum_{n=0}^{\infty}\frac{x^n}{n!}\ensuremath{\ \blacksquare}$

2. Let $f(x) = \sin{x}$. Then since $f^{(n)}(x) = \sin(x + \frac{n\pi}{2})$, we have $f^{(n)}(0) = \sin(\frac{n\pi}{2})$. Now for $k$ is even, $\sin(\frac{k\pi}{2}) = 0$ and for $k = 2m+1$, we have $\sin(\frac{k\pi}{2}) = \sin(\frac{(2m+1)\pi}{2}) = (-1)^m$. Thus

$\displaystyle P(x) = \sum_{k=0}^{n-1}\frac{\sin(\frac{k\pi}{2})}{k!}x^k = \sum_{m=0}^{\frac{n-1}{2}}\frac{(-1)^m}{(2m+1)!}x^{2m+1}. $

Let $b_{2m+1} = \frac{(-1)^m}{(2m+1)!}x^{2m+1}$. Then by the limit ration test,
$\displaystyle \lim_{m \to \infty}\vert\frac{b_{2m+3}}{b_{2m+1}}\vert$ $\displaystyle =$ $\displaystyle \lim_{m\to \infty}\vert\frac{\sin(\frac{(2m+3)\pi}{2}) x^{2m+3}}{(2m+3)!}\cdot \frac{(2m+1)!}{\sin(\frac{(2m+1)\pi}{2}) x^{2m+1}}\vert$  
  $\displaystyle =$ $\displaystyle \lim_{m \to \infty}\frac{x^2}{(2m+3)(2m+2)} = 0 < 1$  

Thus for all $x$, $\sum_{k=0}^{\infty}\frac{\sin(\frac{k\pi}{2})}{k!}x^k$ converges. Therefore, $\lim_{n \to \infty}R_n = 0$ and

$\displaystyle \sin{x} = \sum_{m=0}^{\infty}\frac{(-1)^m}{(2m+1)!}x^{2m+1}\ensuremath{\ \blacksquare}$

Exercise 2..16   Find a MacLaurin series expansion of the following function
1. $f(x) = \frac{1}{1-x}$2. $f(x) = \log(1+x)$

SOLUTION 1. Let $f(x) = \frac{1}{1-x}$. Then since $f^{(n)}(x) = \frac{n!}{(1-x)^{n+1}}$, we have $f^{(n)}(0) = n!$. Thus Taylor polynomial $P(x)$ is $\displaystyle{P(x) = \sum_{k=0}^{n-1}\frac{k!}{k!}x^k = \sum_{k=0}^{n-1}x^k.}$ Let $b_n = \frac{f^{(n)}(0)}{n!}x^n = x^n$. Then by Limit Ratio Test, we have

$\displaystyle \lim_{n \to \infty}\vert\frac{b_{n+1}}{b_n}\vert = \lim_{n\to \infty}\vert\frac{x^{n+1}}{x^n}\vert = \vert x\vert$

Thus for $\vert x\vert < 1$, $\sum_{k=0}^{\infty}x^k$ converges. Therefore, $\lim_{n \to \infty}R_n = 0$ and

$\displaystyle \frac{1}{1-x} = \sum_{n=0}^{\infty}x^n\hskip 0.3cm (\vert x\vert < 1)\ensuremath{\ \blacksquare}$

2. Let $f(x) = \log(1+x)$. Then $f'(x) = \frac{1}{1+x} = (1+x)^{-1}$. Put $t = -x$. Then

$\displaystyle f^{(n+1)}(x) = \frac{d^{n}(\frac{1}{1+x})}{dx^n} = (-1)^n \frac{d^n (\frac{1}{1-t})}{dt^n} = (-1)^n \frac{n!}{(1-t)^n} = \frac{(-1)^n n!}{(1+x)^n}.$

Thus

$\displaystyle f^{(n)}(0) = \left\{\begin{array}{ll} 0 &, n = 0\\ (-1)^{n-1}(n-1)! &, n \geq 1 \end{array}\right.$

Taylor polynomial is given by $P(x) = \sum_{k=0}^{n-1}\frac{(-1)^{k-1}(k-1)!}{k!}x^k = \sum_{k=0}^{n-1}\frac{(-1)^{k-1}}{k}x^k$. Let $b_n = \frac{f^{(n)}(0)}{n!}x^n = \frac{(-1)^{n-1}}{n}x^n$ and apply Limit Ratio Test. Then

$\displaystyle \lim_{n \to \infty}\vert\frac{b_{n+1}}{b_n}\vert = \lim_{n \to \i... ...ac{n}{x^n}\vert = \lim_{n \to \infty}\vert\frac{(n+1)x}{n}\vert = \vert x\vert.$

Thus for $\vert x\vert < 1$, $\sum_{k=0}^{\infty}\frac{(-1)^{k-1}}{k}x^k$ converges. Therefore, $\lim_{n \to \infty}R_n = 0$ and

$\displaystyle \log(1+x) = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}x^n\hskip 0.3cm (-1 < x \leq 1) \ensuremath{\ \blacksquare}$

Landau little o

Theorem 2..17   Suppose a function $f(x)$ has Maclaurin series expansion. Then

$\displaystyle R_n = \frac{f^{(n)}(0)}{n!}{x^n} + o(x^n)$

where, $\lim_{x \to 0}\frac{o(x^n)}{x^n} = 0.$

Proof $\displaystyle{\lim_{n \to 0}\frac{(R_n - \frac{f^{(n)}(0)}{n!}x^n)}{x^n} = \lim_{x \to 0}\frac{f^{(n)}(\theta x) - f^{(n)}(0)}{n!} = 0}$

Example 2..17   Evaluate the following limit

$\displaystyle \lim_{x \rightarrow 0}\frac{\cos{x} - 1}{x^{2}} $

SOLUTION Since the denominator is $x^2$, we find Taylor polynomial of 2nd degree of $f(x) = \cos{x}$.

$\displaystyle \cos{x} = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + o(x^2)$

implies

$\displaystyle \cos{x} = 1 - \frac{1}{2}x^2 + o(x^2).$

Thus,
$\displaystyle \lim_{x \rightarrow 0}\frac{\cos{x} - 1}{x^2}$ $\displaystyle =$ $\displaystyle \lim_{x \rightarrow 0}\left(-\frac{1}{2} - \frac{o(x^2)}{x^2}\right) = - \frac{1}{2} \ensuremath{\ \blacksquare}$  

Exercise 2..17   Evaluate the following limit.

$\displaystyle \lim_{x \rightarrow 0}\frac{\log(1+x) - x + \frac{x^2}{2}}{x^{3}} $

SOLUTION Since the denominator is $x^3$, we find Taylor polynomial of 3rd degree of $f(x) = \log(1+x)$. $f'(x) = \frac{1}{1+x}, f''(x) = -\frac{1}{(1+x)^2}, f'''(x) = \frac{2}{(1+x)^3}$implies ,

$\displaystyle \log(1+x)$ $\displaystyle =$ $\displaystyle f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + \frac{f'''(0)}{3!}x^3 + o(x^3)$  
  $\displaystyle =$ $\displaystyle x - \frac{x^2}{2} + \frac{2x^3}{3!} + o(x^3).$  

Thus,
$\displaystyle \lim_{x \rightarrow 0}\frac{\log(1+x) - x + \frac{x^2}{2}}{x^{3}} $ $\displaystyle =$ $\displaystyle \lim_{x \rightarrow 0}\big(\frac{2}{3!} + \frac{o(x^3)}{x^3}\big) = \frac{1}{3} \ensuremath{\ \blacksquare}$  

Exercise A


1.
Find a MacLaurin series expansion of the following functions?D (a) $\displaystyle{\frac{1}{1+x}}$

(b) $\displaystyle{\frac{1}{1-x^2}}$

(c) $\displaystyle{\frac{1}{\sqrt{1-x^2}}} $

(d) $\displaystyle{\sqrt{1-x^2}} $

Exercise B


1.
Show the following MacLaurin series expansion holds?D

(a) $\displaystyle{\cos{x} = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \cdots + (-1)^{n}\frac{x^{2n}}{(2n)!} + \cdots,\ (-\infty < x < \infty)} $

(b) $\displaystyle{\log(1+x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \cdots + (-1)^{n-1}\frac{x^{n}}{n} + \cdots , \ (-1 < x \leq 1))}$

(c) $\displaystyle{(1+x)^{\alpha} = 1 + \frac{\alpha}{1!}x + \frac{\alpha(\alpha-1)}{2!}x^{2} + \cdots + \frac{\alpha(\alpha-1)\cdots(\alpha -n+1)}{n!}x^{n} + \cdots}$
???????C $(-1 < x < 1)$

(d) $\displaystyle{\tan^{-1}{x} = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \cdots + \frac{(-1)^{n}x^{2n+1}}{2n+1} + \cdots}$, $(-1 < x < 1)$

2.
Find the limit of the following functions using Landau o ?D

(a) $\displaystyle{\lim_{x \rightarrow 0}\frac{\log{(1+x)}}{x}}$ (b) $\displaystyle{\lim_{x \rightarrow 0}\frac{x - \sin{x}}{x^3}}$ (c) $\displaystyle{\lim_{x \rightarrow 0}\frac{e^{x} - 1 - x}{x^2}}$ (d) $\displaystyle{\lim_{x \rightarrow \infty}\frac{x^{\alpha}}{e^{x}}}$

3.

(a) By the exercise 1(d)?Cwe can obtain $\frac{\pi}{4} = \tan^{-1}(1) = 1 - \frac{1}{3} + \frac{1}{5} + \frac{1}{7} - \cdots $ Now using this fact, calculate $\pi$?D

(b) $\frac{\pi}{4} = 4\tan^{-1}(\frac{1}{5}) - \tan^{-1}(\frac{1}{239})$ is called Machin's formula?DUsing this formula, calculate $\pi$ 100 digts after the decimal point?D



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