Maclaurin Polynomials
In this section, expand on the discussion of Taylor polynomials to introduce the concept of Maclaurin Polynomials.
Recall that a Taylor series will be a power series which makes use of various derivatives of a given function, such as first, second, third derivatives.
The Taylor series representation is written as:
\[ \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n = f(a) + f'(a)(x – a) + \frac{f”(a)}{2!} (x – a)^2 + \frac{f”'(a)}{3!} (x – a)^3 + \cdots.\]
This power series above is known as the Taylor series for f at a.
In the special case where a = 0, we refer to the resulting series as a Maclaurin polynomial.
Thus the Maclaurin power series is given by:
\[ \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n = f(0) + f'(0)x + \frac{f”(0)}{2!} x^2 + \frac{f”'(0)}{3!} x^3 + \cdots \]
The video below explores the setup of these Maclaurin series representations and provides several worked out examples.
Note that a Maclaurin polynomial takes a smooth function 饾憮(饾懃) and represents the function as a sum of its derivatives at a single point, multiplied by powers of 饾懃and divided by factorials. Instead of constructing an infinite series, we sometimes build a polynomial with a limited degree of terms, such as 4 terms or 5 terms. If the number of terms in the polynomial is n, then this is referred to as an nth Maclaurin polynomial. As the degree of the polynomial increases, the approximation becomes increasingly accurate within a certain interval around the origin.
Example 1: Maclaurin Polynomial for 饾拞饾挋
Let 饾憮(饾懃) = 饾憭饾懃.
Compute derivatives:
饾憮鈥�(饾懃) = 饾憭饾懃,饾憮鈥测��(饾懃) = 饾憭饾懃,饾憮(饾忆)(饾懃) = 饾憭饾懃
Evaluate at 饾懃 = 0:
饾憮(饾忆)(0) = 1
Thus, the 饾憶th Maclaurin polynomial is:
\[ P_n(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!} \]
Example calculation (n = 4):
\[ P_4(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} \]
Approximation Example:
To approximate 饾憭0.5:
饾憙4(0.5) = 1+0.5+0.125+0.0208+0.0026 = 1.6484
Actual 饾憭0.5 鈮� 1.6487, showing excellent accuracy even with only four terms.
Example 2: Maclaurin Polynomial for 饾惉饾悽饾惂 (饾挋)
Let 饾憮(饾懃) = sin (饾懃).
Compute derivatives:
饾憮鈥�(饾懃) = cos (饾懃),饾憮鈥测��(饾懃) = 鈭抯in (饾懃),饾憮(3)(饾懃) = 鈭抍os (饾懃),饾憮(4)(饾懃) = sin (饾懃)
Evaluate at 饾懃 = 0:
饾憮(0) = 0, 鈥婐潙撯��(0) = 1, 鈥婐潙撯�测��(0) = 0, 鈥婐潙�(3)(0) = 鈭�1, 鈥婐潙�(4)(0) = 0
Substitute into the formula:
\[ P_n(x) = x – \frac{x^3}{3!} + \frac{x^5}{5!} – \frac{x^7}{7!} + \cdots \]
This alternating pattern continues for all odd powers.
Example (n = 5):
\[ P_5(x) = x – \frac{x^3}{6} + \frac{x^5}{120} \]
Small-angle approximation:
For small 饾懃, sin (饾懃) 鈮� 饾懃. This simplification is widely used in physics for small oscillations and pendulum motion.
Example Calculation:
Approximate sin (0.5):
\[ P_5(0.5) = 0.5 – \frac{(0.5)^3}{6} + \frac{(0.5)^5}{120} = 0.4794 \]
Actual sin (0.5) = 0.4794, showing near-perfect accuracy.
Example 3: Maclaurin Polynomial for 饾悳饾惃饾惉 (饾挋)
Let 饾憮(饾懃) = cos (饾懃).
Derivatives:
饾憮鈥�(饾懃) = 鈭抯in (饾懃),饾憮鈥测��(饾懃) = 鈭抍os (饾懃),饾憮(3)(饾懃) = sin (饾懃),饾憮(4)(饾懃) = cos (饾懃)
Evaluate at 饾懃 = 0:
饾憮(0) = 1, 鈥婐潙撯��(0) = 0, 鈥婐潙撯�测��(0) = 鈭�1, 鈥婐潙�(3)(0) = 0, 鈥婐潙�(4)(0) = 1
Substitute:
\[ P_n(x) = 1 – \frac{x^2}{2!} + \frac{x^4}{4!} – \frac{x^6}{6!} + \cdots \]
Example (n = 4):
\[ P_4(x) = 1 – \frac{x^2}{2} + \frac{x^4}{24} \]
Approximation Example:
饾憙4(0.5) = 1 鈭�0.125 +0.0052 = 0.8802
Actual cos (0.5) = 0.8776, again demonstrating excellent agreement.
Example 4: Maclaurin Polynomial for 饾惀饾惂 (饾煆 + 饾挋)
Let 饾憮(饾懃) = ln (1 +饾懃).
Compute derivatives:
\[ f'(x) = \frac{1}{1+x},f”(x) = -\frac{1}{(1+x)^2},f^{(3)}(x) = \frac{2}{(1+x)^3},f^{(4)}(x) = -\frac{6}{(1+x)^4} \]
Evaluate at 饾懃 = 0:
饾憮(0) = 0, 鈥婐潙撯��(0) = 1, 鈥婐潙撯�测��(0) = 鈭�1, 鈥婐潙�(3)(0) = 2, 鈥婐潙�(4)(0) = 鈭�6
Substitute:
\[ P_n(x) = x – \frac{x^2}{2} + \frac{x^3}{3} – \frac{x^4}{4} + \cdots + (-1)^{k+1} \frac{x^k}{k} \]
This is a power series for ln (1 + 饾懃), converging for 鈭� 饾懃 鈭�< 1.
Example (n = 4):
\[ P_4(x) = x – \frac{x^2}{2} + \frac{x^3}{3} – \frac{x^4}{4} \]
Approximate ln (1.2):
饾憙4(0.2) = 0.2 鈭� 0.02 +0.0027 鈭�0.0004 = 0.1823
Actual ln (1.2) = 0.1823. Perfect agreement up to 4 decimals.
Example 5: Maclaurin Polynomial for (饾煆 + 饾挋)饾挀
Let 饾憮(饾懃) = (1 +饾懃)饾憻, where 饾憻 is any real number.
Compute derivatives:
饾憮鈥�(饾懃) = 饾憻(1+饾懃)饾憻鈭�1,饾憮鈥测��(饾懃) = 饾憻(饾憻 鈭�1)(1+饾懃)饾憻鈭�2,鈥�
Evaluate at 饾懃 = 0:
饾憮(饾憳)(0) = 饾憻(饾憻 鈭�1)(饾憻 鈭�2)鈰�(饾憻 鈭掟潙� +1)
Substitute:
\[ P_n(x) = 1 + rx + \frac{r(r-1)}{2!}x^2 + \frac{r(r-1)(r-2)}{3!}x^3 + \cdots \]
This is the generalized binomial expansion.
Example (r = 陆):
\[ (1+x)^{1/2} = 1 + \frac{1}{2}x – \frac{1}{8}x^2 +\frac{1}{16}x^3 – \frac{5}{128}x^4 + \cdots \]
