logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

Obtain the Maclaurin's series expansion for:$\;\cos^{2}x$

Note: This is part 2nd of a 4 part question, split as 4 separate questions here.

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Maclaurin's series :If $ f(x)$ and its derivatives (a sufficiently large number of derivatives ) are continuous on [o x], then Maclaurin series for the expansion of $f(x)$ is given by $f(x)=f(0)+\large\frac{f'(0)}{1!} x+\frac{f''(0)}{2 !}x^2+............$
Given $\cos ^2 x$
Step 1:
$f(x)=\cos ^2 x$                          $f(0)=\cos^2=1$
$f'(x)=-2\cos x \sin x$            $f'(0)=\sin 0=0$
$\qquad=-\sin 2x$
$f''(x)=-\cos 2x$                      $f''(0)=-2$
$f'''(x)=+4\sin 2x$                     $f'''(0)=0$
$f^{iv}(x)=8 \cos 2x$                    $f^{iv} (0)=8$
Step 2:
Maclaurin's series for $f(x)$ is given by
$f(x)=f(0)+\large\frac{f'(0)}{1!} x+\frac{f''(0)}{2 !}x^2+............$
Step 3:
$\cos ^2x=1- \large\frac{2x^2}{2!}+\frac{8x^4}{4 !}+.....$
$\qquad=1-x^2+\large\frac{x^4}{3}+....$

 

answered Jul 25, 2013 by meena.p
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...