Clifland

2021-11-06

The Maclaurin series of f(x) if's first four terms should be written out

Bentley Leach

Expert

Maclaurin series:
$f\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{f}^{\left(n\right)\left(0\right)}}{n!}{x}^{n}=f\left(0\right)+{f}^{\prime }\left(0\right)x+\frac{f{}^{″}\left(0\right)}{2!}{x}^{2}+\frac{f{}^{‴}\left(0\right)}{3!}{x}^{3}+\frac{{f}^{\left(4\right)}\left(0\right)}{4!}{x}^{4}+\dots$
Here given $f\left(0\right)=2,{f}^{\prime }\left(0\right)=3,{f}^{″}\left(0\right)=4$ and we need to find out the first 4 terms of the Maclaurin Series
$f\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{f}^{\left(n\right)}\left(0\right)}{n!}{x}^{n}=f\left(0\right)+{f}^{\prime }\left(0\right)x+\frac{f{}^{″}\left(0\right)}{2!}{x}^{2}+\frac{f{}^{‴}\left(0\right)}{3!}{x}^{3}+\dots$
We need to find 4 terms
$f\left(x\right)=2+3x+\frac{4}{2}{x}^{2}+\frac{12}{6}{x}^{3}$
$f\left(x\right)=2+3x+2{x}^{2}+2{x}^{3}$
Result: $f\left(x\right)=2+3x+2{x}^{2}+2{x}^{3}$

Do you have a similar question?

Recalculate according to your conditions!