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Maclaurin Series Formula
Maclaurin series formula helps in writing a function as a series (or sum) of terms involving the derivatives of the function. This formula helps in finding the approximate value of the function. In math, sometimes it is really difficult to evaluate some functions. In such cases, we use the approximation formulas where the function is expressed as a series. There are two such approximation formulas:
 Taylor series formula
 Maclaurin series formula
Let us learn these along with a few solved examples in the upcoming sections.
What is Maclaurin Series Formula?
The Maclaurin series formula is a special case of the Taylor series formula. The Taylor series of a function f(x) (which is a differentiable function) at x = a is:
\(\begin{align}
f(x) &=\sum_{n=0}^{\infty} \dfrac{f^{(n)}(a)}{n !}(xa)^{n} \\
&=f(a)+f^{\prime}(a)(xa)+\dfrac{f^{\prime \prime}(a)}{2 !}(xa)^{2}+\dfrac{f^{\prime \prime \prime}(a)}{3 !}(xa)^{3}+\cdots
\end{align}\)
Maclaurin series is nothing but the Taylor series when a = 0. i.e., the Maclaurin series formula is obtained by substituting a = 0 in the above formula. Thus, the Maclaurin series formula is,
\(\begin{align}
f(x) &=\sum_{n=0}^{\infty} \dfrac{f^{(n)}(0)}{n !} x^{n} \\
&=f(0)+f^{\prime}(0) x+\dfrac{f^{\prime \prime}(0)}{2 !} x^{2}\\[0.2cm]&+\dfrac{f^{\prime \prime \prime}(0)}{3 !} x^{3}+\cdots
\end{align}\)
Here are Maclaurin series expansions (which can be found by using the above formula) for some commonly used functions.
 \(e^{x}=1+\dfrac{x}{1 !}+\dfrac{x^{2}}{2 !}+\dfrac{x^{3}}{3 !} \cdots=\sum_{n=0}^{\infty} \dfrac{x^{n}}{n !}\)
 \(\sin (x)=x\dfrac{x^{3}}{3 !}+\dfrac{x^{5}}{5 !}\dfrac{x^{7}}{7 !}+\cdots=\sum_{n=0}^{\infty} \dfrac{(1)^{n} x^{2 n+1}}{(2 n+1) !}\)
 \(\cos (x)=1\dfrac{x^{2}}{2 !}+\dfrac{x^{4}}{4 !}\dfrac{x^{6}}{6 !}+\cdots=\sum_{n=0}^{\infty} \dfrac{(1)^{n} x^{2 n}}{(2 n) !}\)
 \(\dfrac{1}{1x}=1+x+x^{2}+x^{3} \cdots=\sum_{n=0}^{\infty} x^{n}\), 1<x<1
 \(\ln (1+x)=x\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3}\dfrac{x^{4}}{4}+\cdots=\sum_{n=0}^{\infty} \dfrac{(1)^{n} x^{n+1}}{n+1}\)
We can see how to derive these expansions using the Maclaurin series formula in the following solved examples section.
Solved Examples Using Maclaurin Series Formula

Example 1: Find the Maclaurin series expansion of the function f(x) = e^{x}.
Solution:
We will find the derivatives of the given function f(x) = e^{x}.
f '(x) = e^{x}
f '' (x) = e^{x}
f ''' (x) = e^{x}
We can clearly see that f(0) = f ' (0) = f '' (0) = f ''' (0) = .... = e^{0} = 1.
Using the Maclurin series formula,
\(f(x) = f(0)+f^{\prime}(0) x+\dfrac{f^{\prime \prime}(0)}{2 !} x^{2}+\dfrac{f^{\prime \prime \prime}(0)}{3 !} x^{3}+\cdots\)
\(= 1+ x+ \dfrac 1 {2!} x^2 + \dfrac 1 {3!} x^3+...\)
Answer: \(e^{x}=\) \(1+\dfrac{x}{1 !}+\dfrac{x^{2}}{2 !}+\dfrac{x^{3}}{3 !} \cdots\) = \(\sum_{n=0}^{\infty} \dfrac{x^{n}}{n !}\)

Example 2 : Find the Maclaurin series expansion of the function f(x) = sin x.
Solution:
We will find the derivatives of the given function f(x) = sin x.
f '(x) = cos x.
f '' (x) =  sin x.
f ''' (x) =  cos x.
f^{(4) }(x) = sin x
f^{(5)} (x) = cos x
We can clearly see that
f(0) = f '' (0) = f '' (0) = f^{(4)}(0) = .... = 0
f ' (0) = 1
f ''' (0) = 1
f^{(5)} (0) = 1
Using the Maclurin series formula,
\(f(x) = f(0)+f^{\prime}(0) x+\dfrac{f^{\prime \prime}(0)}{2 !} x^{2}+\dfrac{f^{\prime \prime \prime}(0)}{3 !} x^{3}+\cdots\)
\(= x\dfrac{x^{3}}{3 !}+\dfrac{x^{5}}{5 !}\dfrac{x^{7}}{7 !}+\cdots\)
Answer: \(\sin x \) = \(x\dfrac{x^{3}}{3 !}+\dfrac{x^{5}}{5 !}\dfrac{x^{7}}{7 !}+\cdots\) = \(\sum_{n=0}^{\infty} \dfrac{(1)^{n} x^{2 n+1}}{(2 n+1) !}\)
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