banner

Write out the first four terms of the maclaurin series of f(x).

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

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

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

5/5 - (18 votes)