Taylor Series and Maclaurin Series

Suppose a real function $f(x)$ has a power series representation about $c$. That is

for some positive real radius of convergence $R$. Then, the coefficients are given by the formula

A power series representation of a function is referred to as the Taylor series. In situations where $c=0$, which occur quite frequently, it’s known as the Maclaurin series. The method to represent a function as an infinite series is called the Taylor expansion.

Proof 1: To obtain $a_0$, we plug $c$ into the function. We differentiate both sides of the functional equation to get

Plugging in $c$, we deduce that $a_1=f’(c)$. We repeat the process to obtain $a_2=\frac{f’‘(c)}{2!}$. By repeating this process $n$ times, we conclude that $a_n=\frac{f^{(n)}(c)}{n!}$.

Proof 2 (Using Integration by Parts):

We have

Now, we use the integration by part. We set the following:

Writing $v$ in this format is special. However, this setup yields the following:

Similarly, we let $u=f’‘(t)$ and $dv=(x-t)dt$, and so we have $du=f^{(3)}(t)dt$ and $v=\frac{-(x-t)^2}{2}$. It implies

Putting all of these together, we have

By repeating this process, we prove the formula.

Motivation of Taylor Expansion

The Taylor Expansion is motivated by the idea of approximating complex functions with polynomials. The underlying concept is that polynomials are easy to work with, and by adding more terms to our series, we can get closer and closer to the actual function.

This allows us to estimate the values of functions, such as trigonometric functions, at angles that are neither special angles nor deducible by trigonometric identities, e.g., $\cos(1)$.

Maclaurin Series of Basic Functions

Use the formula

1. $e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$ (Since the $n$th derivative of $e^x$ is $e^x$, so $C_n=\frac{1}{n!}$ by the formula.)
2. $\sin(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}$ (Observe the derivative of $\sin(x)$, which has period $4$, and we can deduce the Maclaurin series.)
3. Binomial expansion $\displaystyle(1+x)^\alpha=\sum_{n=0}^\infty\binom{\alpha}{n}x^n$ where $\binom{\alpha}{n}=\frac{\alpha(\alpha-1)\cdot(\alpha-2)\cdots(\alpha-n+1)}{n!}$.

Use the known Maclaurin series

1. By taking the derivative of both sides of the sine function and its power series, we obtain the Maclaurin series for $\cos(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}$ .

2. By taking the derivative of $\ln(1+x), $ we have $\frac{1}{1+x}$ which has the Maclaurin series

   \[\frac{1}{1+x}=\sum_{n=0}^\infty(-x)^n\]

   Therefore, the Maclaurin series for $\ln(1+x)$ can be derived by integrating both sides of the above equation, that is

   \[\begin{align*} \ln(1+x)&=\int_0^x\frac{1}{1+t}dt = \int_0^x\sum_{n=0}^\infty(-1)^nt^ndt\\\\ &=\sum_{n=0}^\infty\int_0^x(-1)^nt^ndt = \sum_{n=0}^\infty(-1)^n\frac{t^{n+1}}{n+1}dt \end{align*}\]

3. The derivative of $\arctan(x)$ is $\frac{1}{1+x^2}$, we can use the similar manner to derive the Maclaurin series for $\arctan(x)$. That is

   \[\begin{align*} \arctan(x)&=\int_0^x\frac{1}{1+t^2}dx=\int_0^x\sum_{n=0}^\infty(-t^2)dt\\\\ &=\sum_{n=0}^\infty\int_0^x(-1)^nt^2dt = \sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}. \end{align*}\]

Table of Maclaurin Series of Basic Functions

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