Chapter 8.7, Problem 29E

To determine

To obtain the Maclaurin series for the function: $f(x)=x\cos \left(\frac{x^2}{2}\right)$ .

Answer to Problem 29E

The Maclaurinseries expansion of $x\cos \left(\frac{x^2}{2}\right)=x\left[1-{\left(\frac{1}{2}\right)}^2\left(\frac{x^4}{2!}\right)+{\left(\frac{1}{2}\right)}^4\left(\frac{x^8}{4!}\right)-{\left(\frac{1}{2}\right)}^6\left(\frac{x^{12}}{6!}\right)+\mathrm{.........}\right]$

Explanation of Solution

Given information: $f(x)=x\cos \left(\frac{x^2}{2}\right)$

Formula used: The Maclaurin expansion of the function:

  $\cos x={\displaystyle \sum _{n=0}^{\infty }{(-1)}^n\left(\frac{x^{2n}}{(2n)!}\right)}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\mathrm{........}$

Calculation : Replace $x\text{to}{x}^2/2$ in cosine term only, the above expression becomes as:

  $\begin{array}{l}x\cos \left(\frac{x^2}{2}\right)=x\left[1-\frac{{\left(x^2/2\right)}^2}{2!}+\frac{{\left(x^2/2\right)}^4}{4!}-\frac{{\left(x^2/2\right)}^6}{6!}+\mathrm{.......}\right]\\ =x\left[{\displaystyle \sum _{n=0}^{\infty }{(-1)}^n{\left(\frac{1}{2}\right)}^{2n}\left(\frac{x^{4n}}{(2n)!}\right)}\right]\\ =x\left[1-{\left(\frac{1}{2}\right)}^2\left(\frac{x^4}{2!}\right)+{\left(\frac{1}{2}\right)}^4\left(\frac{x^8}{4!}\right)-{\left(\frac{1}{2}\right)}^6\left(\frac{x^{12}}{6!}\right)+\mathrm{.........}\right]\\ =x\left[{\displaystyle \sum _{n=0}^{\infty }{(-1)}^n\left(\frac{{\left(x^2/2\right)}^{2n}}{(2n)!}\right)}\right]\end{array}$

Hence, the Maclaurin expansion of the function:

  $x\cos \left(\frac{x^2}{2}\right)=x\left[1-{\left(\frac{1}{2}\right)}^2\left(\frac{x^4}{2!}\right)+{\left(\frac{1}{2}\right)}^4\left(\frac{x^8}{4!}\right)-{\left(\frac{1}{2}\right)}^6\left(\frac{x^{12}}{6!}\right)+\mathrm{.........}\right]$