10-31-25 Taylor and Maclaurin Polynomials

Q1.

$f(x) = \frac 1{(3-x)²}$

$=(3-x)^{-2} \Rightarrow (3(1-\frac x3))^{-2}$

$=3^{-2} (1-\frac x3)^{-3}$

$=\frac 19 (1+ ( \frac {-x}3) ^{-2}$

$=\frac 19 ( \sum_{n=0}^\infty (^{-2}_n) (\frac {-x}3)^n)$

Binomial Series

Tee binomial expansion for $f(x) = (1 + x)^k$ is

$(1+x)^k = \sum_{n=0}^k (^k_n) x^n$

where k is a positive integer greater than or equal to n, and the binomial coefficient is

$(^k_n) = \frac {k(k-1)(k-2)…(k-n+1)}{n!}$

While this formula for the binomial coefficient is usually used when k an n are nonnegative integers, we use this formula as our definition of $(^k_n)$ for any real number k and nonnegative integer n.

Theorem 2. Algebra of Power Series

Let f(x) = \sum_{n=0}^\infty a_nx^n and g(x) = \sum_{n=0}^\infty b_nx^n with a common interval of convergence I. Then:

• 1. The power series for $f$ and $g$ can be added or subtracted to obtain a newer power series for their sum or difference with interval of convergence at least as large as I:

(f \pm g)(x) = f(x) \pm g(x) = \sum_{n=0}^\infty a_nx^n \pm \sum_{n=0}^\infty b_nx^n = \sum_{n=0}^\infty (a_n \pm b_n)x^n

• 2. The power series for $f$ and $g$ can be multiplied to obtain a new power series for their product with interval of convergence at least as large as I:

$(f \cdot g)(x) = f(x) \cdot g(x) = (\sum_{n=0}^\infty a_nx^n)(\sum_{n=0}^\infty b_nx^n)$

$= a_0b_0 + (a_0b_1 + a_1b_0)x + (a_0b_2+a_1b_1+a_2b_0)x² + …$

• 3. The power series for $f$ and $g$ can be divided provided $b_0 \ne 0$ to obtain a new power series for their quotient. The interval of convergence is generally difficult to determine

Example 57. Evaluate $\lim_{x\to 0} \frac {\cos x - 1 + \frac {x²}2}{x^4}$ using the Maclaurin series for $\cos x$

Solution: 

$\lim_{x\to 0} \frac {\cos x - 1 + \frac {x²}2}{x^4} \lim_{x\to 0} \frac {(1-\frac {x²}{2!} + \frac {x^4}{4!}-\frac {x^6}{6!}+…)-1 + \frac {x²}{2}}{x^4}$

Since we can add and subtract convergent power series, we obtain

$=\lim_{x\to 0} (\frac {1}{4!} - \frac {x²}{6!} + \frac {x^4}{8!} - …)$

$=\frac 1{24}$

Taylor Polynomials and Remainders

When a Taylor series for $f$ convergent for all $x$ in an interval I, by definition it is also true that the sequence of partial sums {$T_n(x)$} associated with the Taylor series for $f(x)$ also converges. 

$T_n(x) = f(a) + \frac {f’(a)}{1!}(x-a) + \frac {f’’(a)}{2!}(x-a)² + … + \frac {f^{(n)} (a)}{n!} (x-a)^n$

Notice that each $T_n(x)$ is a sum of only a finite number of terms. It is a polynomial of degree at most n and is called the $n^{th}-$ degree Taylor polynomial for $f(x)$ centered at $x=a$. If $a=0$ then the polynomial is called the $n^{th}-$ degree Maclaurin polynomial for $f(x)$ centered at $x=0$.

Theorem 3. Taylor Series Representation of Functions

Let $f$ be a function which has a Taylor series about a. Let$T_n(x)$ be the $n^{th}-$ degree Taylor polynomial of f, and let $R_n(x) = f(x) - T_n(x).$ If r>0 and $\lim_{n\to\infty} R_n(x) = 0$ for all $x \epsilon [a-r,a+r]$, then

f(x) = \lim_{n\to\infty} T_n(x) = \sum_{n=0}^\infty \frac {f^{(n)}(a)(x-a)^n} {n!}

on $[a-r,a+r]$. Hence, $f(x)$ is represented by its Taylor series about a on the interval $[a-r,a+r]$. 

The work is to show that $\lim_{n\to\infty} R_n(x) = 0$; to do so, the following inequality is usually used.

Theorem 4. Taylor’s Inequality

Let M>0 and r>0. If $|f^{(n+!)} (x) | \le M$ for $|x-a| \le r$, then the remainder $R_n (x)$ of the Taylor series satisfies the inequality

$|R_n(x)| \le \frac {M} {(n+1)!} |x-a|^{n+1}$ for all x such that $|x-a| \le r$

Example 60. Find the third degree Maclaurin polynomial $T_3(x)$ of $f(x) = e^x$. Use $T_3(1)$ to approximate the value of $e^1$, and use the remainder to estimate the error in this approximation

Solution: Since

e^x =_{approx} T_3 (x) =  \sum_{n=0}³ \frac {x^n} {n!} = 1 + \frac x{1!} + \frac {x²}{2!} + \frac {x³}{3!}

then

e^1 =_{approx} T_3 (1) = \sum_{n=0}³  \frac 1{n!} = \sum_{n=0}^\infty \frac 1{n!} + 1 + 1 + \frac 1{2!} + \frac 1{3!} = 2 \frac 23

The 4th derivative of $f(x)$ is $e^x$, and hence an upper bound for the 4th derivative on the interval $[0, 1]$ is f^{(4)} (1) = e^1 < 3. Using the Taylor inequality we have:

$|R_3 (x)| \le \frac 3{4!} |x-0|^4$    for all $x \epsilon [0,1]$

Hence $|R_3 (1)| \le \frac 3{4!} = \frac 18 = 0.125$. Putting these together we can determine that the exact value of $e^1$ is between T_3 (1) - |R_3(1) - |R_3 (1)| and $T_3(1) + |R_3(1)|$. Hence

$e^1 \epsilon (2\frac 23 - \frac 18, 2 \frac 23 + \frac 18) = (\frac {61}{24}, \frac {67}{24}) =_{approx} (2.54, 2.79)$

Example 63. Consider the function $f(x) = \sin x$

$a)$  Find the Maclaurin Polynomials $T_1 (x), T_3 (x)$ and for f(x) = \sin x and plot all 4 function on the same plot on the interval $[-\pi, \pi]$

$b)$ Calculate $T_1 (\frac {\pi}6), T_3 (\frac {\pi}6),$ and $T_5 (\frac {\pi}6)$ and compare them to $\sin \frac {\pi} 6$

$c)$ Use $T_5 (x)$ to approximate $\sin 8\degree$