I'm not acing Calculus II and while this problem might be easy to some, it's not to me. Any help will do.
Problem: Find the first nonzero terms of the Taylor Series for $f(x) = \cos x$ where $a = {\pi\over 6}$
My work:
\begin{array}{rlrll} f(x) =& \cos x & f(a) =& \cos {\pi\over 6} &= {\sqrt 3\over 2} \\ f'(x) =& - \sin x & f'(a) =& - \sin {\pi\over 6} &= -{1\over 2} \\ f ''(x) =& - \cos x & f ''(a) =& - \cos {\pi\over 6} &= -{\sqrt 3\over 2} \\ f ''' (x) =& - (-\sin x) = \sin x & f ''' (a) =& \sin {\pi\over 6} &= {1\over 2} \\ f^4(x) =& \cos x & f^4(a) =& \cos {\pi\over 6} &= {\sqrt 3\over 2} \end{array}
Taylor Series:
$a_n={f^n(a)\over n!}(x-a)^n$
$\eqalign{ & \frac{{{f^0}(a)}}{{0!}}{(x - a)^0} + \frac{{f'(a)}}{{1!}}{(x - a)^1} + \frac{{f''(a)}}{{2!}}{(x - a)^2} + \frac{{f'''(a)}}{{3!}}{(x - a)^3} + \frac{{{f^4}(a)}}{{4!}}{(x - a)^4} \cr & = \frac{1}{{0!}}\;\frac{{\sqrt 3 }}{2}(1) + \left( { - \frac{1}{2}} \right)\frac{1}{{1!}}\left( {x - \frac{\pi }{6}} \right) + \left( { - \frac{{\sqrt 3 }}{2}} \right)\frac{1}{{2!}}{\left( {x - \frac{\pi }{6}} \right)^2} + \frac{1}{{3!}}\frac{1}{2}{\left( {x - \frac{\pi }{6}} \right)^3} + \frac{1}{{4!}}\frac{{\sqrt 3 }}{2}{\left( {x - \frac{\pi }{6}} \right)^4} \cr & = \;\frac{{\sqrt 3 }}{2} - \frac{1}{2}\left( {x - \frac{\pi }{6}} \right) - \frac{{\sqrt 3 }}{2}\frac{1}{2}{\left( {x - \frac{\pi }{6}} \right)^2} + \frac{1}{2}\frac{1}{{3 \cdot 2}}{\left( {x - \frac{\pi }{6}} \right)^3} + \frac{{\sqrt 3 }}{2}\frac{1}{{4 \cdot 3 \cdot 2}}{\left( {x - \frac{\pi }{6}} \right)^4} \cr & = \;\frac{{\sqrt 3 }}{2} - \frac{1}{2}\left( {x - \frac{\pi }{6}} \right) - \frac{{\sqrt 3 }}{4}{\left( {x - \frac{\pi }{6}} \right)^2} + \frac{1}{{12}}{\left( {x - \frac{\pi }{6}} \right)^3} + \frac{{\sqrt 3 }}{{48}}{\left( {x - \frac{\pi }{6}} \right)^4} \cr} $
Is this problem solved correctly? If not, where did I go wrong? If you have any tips for the Taylor Series I'll take it also. Thank you.