This question aims to find the first four terms of the Maclaurin series when the values of **f(0), f’(0), f’’(0)** and** f’’’(0)** are given.

Maclaurin series is an expansion of **the Taylor series**. It calculates the value of a function f(x) **close to zero**. The value of **successive derivatives** of the function f(x) must be known. The formula for **Maclaurin Series** is given as:

\[\sum_ {n=0}^ {\infty} \dfrac{ f^{n} (a) }{ n! } (x – a)^n \]

## Expert Answer

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

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

\[ f ( x ) = f ( 0 ) + f’ ( 0 ) x + \frac { f’’ ( 0 ) } { 2! } x^2 + \frac { f’’’ ( 0 ) } { 3! } x^3 + \frac { f ^ {(4)} ( 0 ) } { 4! } x^4 + … \]

To find the first four terms of Maclaurin’s series:

\[ f ( x ) = f ( 0 ) + f’ ( 0 ) x + \frac { f’’ ( 0 ) } { 2! } x^2 + \frac { f’’’ ( 0 ) } { 3! } x^3 + … \]

The values of f ( 0 ), f’ ( 0 ), and f’’ ( 0 ) are given so we need to put these values in the above-mentioned series.

These values are:

**f ( 0 ) = 2 , f’ ( 0 ) = 3, f’’ ( 0 ) = 4, f’’’ ( 0 ) = 12**

Putting these values:

\[ f ( x ) = 2 + 3 x + \frac {4}{2} x ^ 2 + \frac {12}{6} x^3 \]

\[ f ( x ) = 2 + 3 x + 2 x ^ 2 + 2 x ^ 3 \]

## Numerical Result

The first four terms of Maclaurin’s series are:

\**[ f ( x ) = 2 + 3 x + 2 x ^ 2 + 2 x ^ 3 \]**

## Example

Find the first two terms of Maclaurin’s series.

\[ f ( x ) = f ( 0 ) + f’ ( 0 ) x + \frac {f’’ ( 0 )}{2!} x^2 + \frac {f’’’ ( 0 )}{3!} x^3 + \frac {f ^ {(4)} ( 0 )}{4!} x^4 + … \]

\[ f ( x ) = f ( 0 ) + f’ ( 0 ) x + \frac{ f’’( 0 ) }{ 2! } x^2 + … \]

The values of f (0) and f’ (0) are given, and they are as follows:

**f ( 0 ) = 4 , f’ ( 0 ) = 2, f’’ ( 0 ) = 6**

\[ f ( x ) = 4 + 2 x + \frac { 6 }{ 2 } x ^ 2 \]

\[ f ( x ) = 4 + 2 x + 3 x ^ 2 \]