Taylor Series

Let $f$ be infinitely many times differentiable on $(a,b)$ and $c,x \in (a,b)$.

If $\lim_{n\to \infty} R_n(x) = 0$ for $x \in (c-R, c+R) \subset (a,b)$, then Taylor series of $f$ at $c$ is given by:

$$\sum_{n=0}^\infty \frac{f^{(n)}(c)}{n!} (x-c)^n$$

Note

Usually Taylor series expansion is done with $c=0$. This is a special case of Taylor series, and called the Maclaurin series. $$

Procedure

Suppose a function $f$ is given and its Taylor series is required. $$

• Differentiate $f$ repeatedly and find a general solution for $n$-th derivative
• Construct Taylor polynomial
• Use root test or ratio test to find the range of convergence of the Taylor polynomial
• Consider the endpoints of range of convergence to check if the Taylor polynomial converges
• Construct the Taylor remainder
• Find for which values of $x$, the remainder converges to $0$

Examples

e^x

Range of convergence is $\mathbb{R}$. $$

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

ln (1+x)

Range of convergence is $(-1,1]$. $$

$$\ln(1+x) = \sum_{n=1}^\infty \frac{(-1)^{n-1}x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots$$

sin x

Range of convergence is $\mathbb{R}$. $$

$$\sin x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$$

cos x

Range of convergence is $\mathbb{R}$. $$

$$\cos x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$$