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 𝑓(𝑥) and represents the function as a sum of its derivatives at a single point, multiplied by powers of 𝑥and 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 𝒆^𝒙

Let 𝑓(𝑥) = 𝑒^𝑥.
Compute derivatives:
𝑓′(𝑥) = 𝑒^𝑥,𝑓′′(𝑥) = 𝑒^𝑥,𝑓^(𝑛)(𝑥) = 𝑒^𝑥

Evaluate at 𝑥 = 0:
𝑓^(𝑛)(0) = 1

Thus, the 𝑛^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 𝑒^0.5:
𝑃₄(0.5) = 1+0.5+0.125+0.0208+0.0026 = 1.6484

Actual 𝑒^0.5 ≈ 1.6487, showing excellent accuracy even with only four terms.

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Example 2: Maclaurin Polynomial for 𝐬𝐢𝐧 (𝒙)

Let 𝑓(𝑥) = sin (𝑥).
Compute derivatives:
𝑓′(𝑥) = cos (𝑥),𝑓′′(𝑥) = −sin (𝑥),𝑓⁽³⁾(𝑥) = −cos (𝑥),𝑓⁽⁴⁾(𝑥) = sin (𝑥)

Evaluate at 𝑥 = 0:
𝑓(0) = 0,  𝑓′(0) = 1,  𝑓′′(0) = 0,  𝑓⁽³⁾(0) = −1,  𝑓⁽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 𝑥, sin (𝑥) ≈ 𝑥. 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.

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Example 3: Maclaurin Polynomial for 𝐜𝐨𝐬 (𝒙)

Let 𝑓(𝑥) = cos (𝑥).
Derivatives:
𝑓′(𝑥) = −sin (𝑥),𝑓′′(𝑥) = −cos (𝑥),𝑓⁽³⁾(𝑥) = sin (𝑥),𝑓⁽⁴⁾(𝑥) = cos (𝑥)

Evaluate at 𝑥 = 0:
𝑓(0) = 1,  𝑓′(0) = 0,  𝑓′′(0) = −1,  𝑓⁽³⁾(0) = 0,  𝑓⁽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:

𝑃₄(0.5) = 1 −0.125 +0.0052 = 0.8802

Actual cos (0.5) = 0.8776, again demonstrating excellent agreement.

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Example 4: Maclaurin Polynomial for 𝐥𝐧 (𝟏 + 𝒙)

Let 𝑓(𝑥) = ln (1 +𝑥).
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 𝑥 = 0:
𝑓(0) = 0,  𝑓′(0) = 1,  𝑓′′(0) = −1,  𝑓⁽³⁾(0) = 2,  𝑓⁽⁴⁾(0) = −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 + 𝑥), converging for ∣ 𝑥 ∣< 1.

Example (n = 4):

\[ P_4(x) = x – \frac{x^2}{2} + \frac{x^3}{3} – \frac{x^4}{4} \]

Approximate ln (1.2):
𝑃₄(0.2) = 0.2 − 0.02 +0.0027 −0.0004 = 0.1823

Actual ln (1.2) = 0.1823. Perfect agreement up to 4 decimals.

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Example 5: Maclaurin Polynomial for (𝟏 + 𝒙)𝒓

Let 𝑓(𝑥) = (1 +𝑥)^𝑟, where 𝑟 is any real number.

Compute derivatives:
𝑓′(𝑥) = 𝑟(1+𝑥)^𝑟−1,𝑓′′(𝑥) = 𝑟(𝑟 −1)(1+𝑥)^𝑟−2,…

Evaluate at 𝑥 = 0:
𝑓^(𝑘)(0) = 𝑟(𝑟 −1)(𝑟 −2)⋯(𝑟 −𝑘 +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 = ½):

\[ (1+x)^{1/2} = 1 + \frac{1}{2}x – \frac{1}{8}x^2 +\frac{1}{16}x^3 – \frac{5}{128}x^4 + \cdots \]