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. LetT_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