Maclaurin Series for Key Functions

Taylor series centered at zero

A Maclaurin series is a Taylor series centered at $0$. This center is especially convenient because powers of $x$ remain unshifted and many elementary functions have simple derivative patterns there. A Maclaurin series is not a list of formulas to recite; each coefficient is determined by derivative values at zero. Knowing the pattern helps, but reconstructing it from derivatives is the safeguard when memory is uncertain.

For the exponential function, every derivative of $e^x$ is $e^x$ and every derivative value at zero is $1$. Therefore

$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots =\sum_{n=0}^{\infty}\frac{x^n}{n!}.$$

The factorial grows rapidly enough that this series converges for every real $x$. Its terms are all positive for positive $x$, so partial sums underestimate $e^x$ there. For negative $x$, signs alternate but the same series remains valid. The formula describes an exact infinite representation, while any finite truncation is an approximation whose error still needs consideration.

It is useful to separate “the series converges” from “a low-degree polynomial is accurate.” The exponential series converges for every input, but a three-term approximation at a large positive input may still omit substantial later terms. Convergence promises that sufficiently many terms approach the correct value; it does not prescribe how many are sufficient for a particular tolerance. A remainder estimate or a numerical comparison supplies that local accuracy decision.

Sine and cosine reveal parity

The derivative pattern for sine repeats every four derivatives:

$$\sin x,\ \cos x,\ -\sin x,\ -\cos x,\ldots$$

At zero, the sine derivatives alternate between $0$, $1$, $0$, and $-1$. Only odd powers survive:

$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots.$$

For cosine, the values at zero are $1$, $0$, $-1$, $0$, and so on, so only even powers survive:

$$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots.$$

This pattern reflects symmetry. Sine is an odd function, meaning $sin(-x)=-\sin x$, so its series contains odd powers. Cosine is even, meaning $\cos(-x)=\cos x$, so its series contains even powers. Symmetry is a check on a proposed series: an even function represented by an unshifted odd-power term has a structural error.

The geometric series generates logarithms and more

The geometric identity

$$\frac{1}{1-x}=1+x+x^2+x^3+\cdots,\qquad |x|<1,$$

is a generative starting point. Integrating it from $0$ to $x$ gives

$$\ln(1-x)=-\left(x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots\right), \qquad -1\le x<1.$$

Substituting $-x$ gives

$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots, \qquad -1<x\le1.$$

The intervals differ at their endpoints, so a sign substitution changes more than appearance. The interior radius remains one, but the numerical endpoint series must be retested after the substitution. This is a concrete reminder that a formula’s interval is part of the formula.

Deriving a familiar series from the geometric series also provides a way to audit signs. Differentiate the logarithm series for $\ln(1+x)$ on its interior interval. The derivative should return $1/(1+x)$, which is the geometric series with $x$ replaced by $-x$. If a proposed coefficient pattern fails this derivative check, it cannot be the correct logarithm representation. Differentiation and integration are therefore not only ways to create series; they are tools for checking them.

Substitute only after preserving structure

To find a series for $e^{-2x}$, substitute $-2x$ for $x$ in the known exponential series:

$$e^{-2x}=1-2x+\frac{(2x)^2}{2!}-\frac{(2x)^3}{3!}+\cdots.$$

The parentheses matter. The power applies to the whole substituted expression, so the coefficient of $x^n$ is $(-2)^n/n!$, not merely $-2/n!$. Write several terms before compressing the result into summation notation. This catches sign and exponent errors that become difficult to see after indexing.

Likewise, substituting $x^2$ into the sine series gives $\sin(x^2)=x^2-x^6/3!+x^{10}/5!-\cdots$. The original odd exponents become exponents congruent to $2$ modulo $4$, because $(x^2)^{2k+1}=x^{4k+2}$. There is no need to derive every new function by repeated differentiation if a valid known series can be transformed transparently.

For any substitution, translate the old interval through the new input before simplifying. The exponential series has infinite radius, so $e^{-2x}$ remains valid for all real $x$. The logarithm series for $\ln(1+u)$ has $-1<u\le1$; putting $u=x^2$ gives $-1<x^2\le1$, which reduces to $-1\le x\le1$ before any endpoint behavior is reconsidered. The lower inequality is automatic because $x^2$ is nonnegative, but the upper inequality is the real restriction. Writing this chain avoids treating interval transformations as an afterthought.

Retrieval check

Write the degree-five Maclaurin polynomial for $\cos x$ and use it to approximate $\cos(0.2)$. Then obtain a series for $\ln(1+x^2)$ from the logarithm series, state its interval of convergence, and identify its first three nonzero terms. Explain why replacing $x$ with $x^2$ requires checking the original condition $|x|<1$ in its new form. The next lesson turns approximation into a controlled claim by bounding the remainder that a finite polynomial leaves behind.