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. transcendental function Let $A_{0}(x),A_{1}(x),\ldots,A_{n}(x)$ be polynomials in $x$. Then a function $y = f(x)$ satisfies a polynomial equation

$$A_{0}(x)y^{n} + \cdots + A_{n}(x) = 0$$

is called algebraic function. A transcendental function is a function that does not satisfy a polynomial equation whose coefficients are themselves roots of polynomials.

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

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

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

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

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

Let

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

Then $F(a) = 0, F(b) = 0$. Thus $F(x)$ satisfies the condition of Rolle's Theorem . Thus

Check

$F(b) = f(b) - f(b) - \sum_{k=1}^{n-1}\frac{f^{(k)}(b)}{k!}(b-b)^{k} - \frac{(b-b)^{n}}{n!}K = 0$.

$$F^{\prime}(\xi) = -f^{\prime}(\xi) - \sum_{k=2}^{n}\frac{f^{(k)}(\xi)}{(k-1)!}(b-\xi)^{k-1} + \sum_{k=1}^{n-1}\frac{f^{(k)}(\xi)}{(k-1)!}(b-\xi)^{k-1}$$

$$+ \frac{(b-\xi)^{n-1}}{(n-1)!}K$$

$$= -f^{\prime}(\xi) + f^{\prime}(\xi) - \frac{f^{n}(\xi)}{(n-1)!}(b-\xi)^{n-1} + \frac{(b-\xi)^{n-1}}{(n-1)!}K = 0$$

Therefore,

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

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

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

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

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

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

Understanding 1. Taylor polynomial around $a = 0$
P(x) = $\sum_{k=0}^{n-1}\frac{f^{(k)}(0)}{k!}x^k$.
2. Lagrange's remainder
$R_n = \frac{f^{(n)}(\theta x)}{n!}$

$\xi \to \theta x$ Note that $\xi \in (0,x)$. Now we show $\xi = \theta x$, where $0 < \theta < 1$. Note that $\theta x = (x - 0)\theta$ can represents all numbers form 0 to $\theta x$. Since $\theta x < x$, we have $0 < \theta < 1$.

Figure 2.8:

Error Estimate Error estimate is to find the bound for the absolute value of remainder term.

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

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

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

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

We next find error estimate. Since

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

we have

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

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

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

$\sin,\cos$ $\cos(x+\pi/2) = -\sin{x} = (\cos{x})'$

Figure 2.9:

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

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

We next find error estimate.

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

Thus error estimate is

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

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

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

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

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

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

$\sum_{k=0}^{n-1} = \sum_{m=0}^{\frac{n-1}{2}}$

Note that $k$ takes values from $k = 0$ to $k = n-1$. Suppose $k = 2m$. Then for $0 = k = 2m$ implies $m =0$ and for $n-1 = k = 2m$ implies $m = \frac{n-1}{2}$.

Understanding In MacLaurin's theorem, a function is expressed by Taylor polynomial and error term. If the size of error estimate is getting close to 0, then we should be able to write $f(x)$ as infinite series.

NOTE

Check

$\sum_{n=0}^{\infty}b_n = \sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{k!}x^k = \sum_{k=0}^{n-1}\frac{f^{(0)}}{k!}x^k + R_n$.

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.

Radius of Convergence Replace $a_n x^n$ by $b_n$. Then $\sum_{n=0}^{\infty}a_n x^n = \sum_{n=0}^{\infty}b_n$. Now $\lim_{n \to \infty}\frac{b_{n+1}}{b_{n}} = \lim_{n \to \infty}\frac{a_{n+1}x^{n+1}}{a_{n}x^n} = \lim_{n \to \infty}\frac{a_{n+1}}{a_{n}}x$. Thus $\lim_{n \to \infty}\left\vert\frac{b_{n+1}}{b_{n}}\right\vert < 1$ implies $\lim_{n \to \infty}\frac{a_{n+1}}{a_{n}}\vert x\vert < 1$ and for $\vert x\vert < \frac{a_n}{a_{n+1}}$, $\sum_{n=0}^{\infty}a_n x^n$ 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

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

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

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

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

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

$\sin{\frac{(2n+1)\pi}{2}} = (-1)^n$

$$\begin{array}{\vert c\vert c\vert c\vert}\hline n & \sin{\fra... ...\ \hline 3 & -1 & -1 \ \hline 4 & 1 & 1 \ \hline \end{array}$$.

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

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

Let $b_{2m+1} = \frac{(-1)^m}{(2m+1)!}x^{2m+1}$. Then by the limit ration test,

$$\lim_{m \to \infty}\vert\frac{b_{2m+3}}{b_{2m+1}}\vert = \lim_{m\to \infty}\vert\frac{\sin(\frac{(2m+3)\pi}{2}) x^{2m+3}}{(2m+3)!}\cdot \frac{(2m+1)!}{\sin(\frac{(2m+1)\pi}{2}) x^{2m+1}}\vert$$

$$= \lim_{m \to \infty}\frac{x^2}{(2m+3)(2m+2)} = 0 < 1$$

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

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

Check

Limit Ratio Test handles consecutive term. But for $b_{2m} = 0$, we need to consider $\frac{2m+3}{2m+2}\cdot \frac{2m+2}{2m+1} = \frac{2m+3}{2m+1}$.

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 $${P(x) = \sum_{k=0}^{n-1}\frac{k!}{k!}x^k = \sum_{k=0}^{n-1}x^k.}$$ Let $b_n = \frac{f^{(n)}(0)}{n!}x^n = x^n$. Then by Limit Ratio Test, we have

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

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

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

Radius of Convergence If $\lim_{n\to \infty}\vert\frac{x^{n+1}}{x^n}\vert = \vert x\vert < 1$, then $\sum_{n=0}^{\infty}x^n$converges. Thus we say $\vert x\vert < 1$ a radius of convergence.

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

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

Thus

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

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

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

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

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

Landau little o

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

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

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

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

Example 2..17   Evaluate the following limit

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

$\sum\frac{(-1)^{n-1}}{n}$

Show $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots$ converges.

$s_{2m+1} = (1 - \frac{1}{2}) + \cdots + (\frac{1}{2m-1} - \frac{1}{2m}) + \frac{1}{2m+1}$
$s_{2m+3} = s_{2m+1} - (\frac{1}{2m+2} - \frac{1}{2m+3})$
Thus $s_{2m+3} < s_{2m+1}$. Then a sequence $s_{2m+1}$ is decreasing and bouded blow by 0. Thus, $s_{2m+1} \to l$. On the other hand, since $s_{2m+2} = s_{2m+1} - \frac{1}{2m+2}$, $s_{2m+2} \to l$. Therefore, $\sum\frac{(-1)^{n-1}}{n}$ converges.

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

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

implies

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

Thus,

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

Exercise 2..17   Evaluate the following limit.

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

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 ,

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

$$= x - \frac{x^2}{2} + \frac{2x^3}{3!} + o(x^3).$$

Thus,

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

Exercise A

1.

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

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

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

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

Exercise B

1.

Show the following MacLaurin series expansion holds

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

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

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

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

2.

Find the limit of the following functions using Landau o

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

3.

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

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