Maclaurin Series Formula

Maclaurin Series Formula

A Maclaurin series is a function that has expansion series that gives the sum of derivatives of that function. The Maclaurin series of a function

\(\begin{array}{l}f(x)\end{array} \)
up to order n may be found using Series
\(\begin{array}{l}[f,  {x, 0, n}]\end{array} \)

It is a special case of Taylor series when x = 0. The Maclaurin series is given by

\[\large f(x)=f(x_{0})+{f}'(x_{0})(x-x_{0})+\frac{{f}”(x_{0})}{2!}(x-x_{0})^{2}+\frac{{f}”'(x_{0})}{3!}(x-x_{0})^{3}+…..\]

The Maclaurin series formula is

\[\large f(x)=\sum_{n=0}^{\infty}\frac{f^{n}(x_{0})}{n!}(x-x_{0})\]

Where,
f(xo), f’(xo), f’‘(xo)……. are the successive differentials when xo = 0.

Function Maclaurin Series
 
\(\begin{array}{l}e^{x}\end{array} \)
 
\(\begin{array}{l}\sum_{k=0}^{\infty}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+…..\end{array} \)
 
\(\begin{array}{l}sin\;x\end{array} \)
 
\(\begin{array}{l}\sum_{k=0}^{\infty}(-1)^{2}=\frac{x^{2k+1}}{(2k+1)!}=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}+\frac{x^{7}}{7!}+…..\end{array} \)
 
\(\begin{array}{l}cos\;x\end{array} \)
\(\begin{array}{l}\cos x=\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2 n}}{(2 n) !}=1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\frac{x^{6}}{6 !}+\ldots\end{array} \)
 
\(\begin{array}{l}\frac{1}{1-x}\end{array} \)
 
\(\begin{array}{l}\sum_{k=0}^{\infty}x^{k}=1+x+x^{2}+x^{3}+….(if-1<x<1)\end{array} \)
  
\(\begin{array}{l}ln(1+x)\end{array} \)
\(\begin{array}{l}\ln (1+x)=\sum_{n=1}^{\infty}(-1)^{n+1} \frac{x^{n}}{n}=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\cdots\end{array} \)

Solved Examples

Question 1: Expanding

\(\begin{array}{l}e^{x}\end{array} \)
: Find the Maclaurin Series expansion of
\(\begin{array}{l}f(x)=e^{x}\end{array} \)

Solution:

Recalling that the derivative of the exponential function is 

\(\begin{array}{l}{f}'(x)=e^{x}\end{array} \)
 In fact, all the derivatives are
\(\begin{array}{l}e^{x}\end{array} \)
.

\(\begin{array}{l}{f}'(0)=e^{0}=1\end{array} \)

\(\begin{array}{l}{f}”(0)=e^{0}=1\end{array} \)

\(\begin{array}{l}{f}”'(0)=e^{0}=1\end{array} \)

We see that all the derivatives, when evaluated at x = 0, give us the value 1.

Also, f(0)=1, so we can conclude the Maclaurin Series expansion will be simply:

\(\begin{array}{l}e^{x}\approx 1+x+\frac{1}{2}x^{2}+\frac{1}{6}x^{3}+\frac{1}{24}x^{4}+\frac{1}{120}x^{5}+….\end{array} \)

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