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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.

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  • $\begingroup$ I didn't check all the arithmetic, but it looks correct. Note the first four non-zero terms would go to the third power, you have one extra term that wasn't asked for. Also, don't write, for example, $f(x)=\cos{\pi/6}$. This isn't true, but $f(\pi/6)=\cos{\pi\over6}$. $\endgroup$ – David Mitra Apr 10 '12 at 22:51
  • $\begingroup$ thanks for the tip. it makes sense now that you mention it, that f(pi/6) would be cos pi/6 because f(a) = cos pi/6, right? $\endgroup$ – fragilewindows Apr 10 '12 at 22:54
  • $\begingroup$ $f(x)=\cos(x)$, you mean. I was referring to the right hand side of your table, where you consistently missused notation ($f^4(x)=\cos{\pi\over 6}$ at the bottom e.g.) $\endgroup$ – David Mitra Apr 10 '12 at 22:55
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    $\begingroup$ To verify your answer, just type taylor cos x around a = pi/6 into WolframAlpha. $\endgroup$ – TMM Jul 11 '12 at 22:49
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Recall that $f(h+a)=\cos(h+(\pi/6))=\cos(h)\cos(\pi/6)-\sin(h)\sin(\pi/6)$. Hence, $$2f(h+a)=\sqrt3\cos(h)-\sin(h). $$ Now, $\cos(h)=1-\frac12h^2+O(h^4)$ and $\sin(h)=h-\frac16h^3+O(h^4)$, hence $$ 2f(h+a)=\sqrt3(1-\tfrac12h^2)-(h-\tfrac16h^3)+O(h^4). $$ The first four terms of the expansion of $f(h+(\pi/6))$ are the $1$, $h$, $h^2$ and $h^3$ terms, namely, $$ f\left(h+\frac\pi6\right)=\frac{\sqrt3}2-\frac12h-\frac{\sqrt3}4h^2+\frac1{12}h^3+O(h^4). $$ More generally, the $h^{2n}$ term of the Taylor expansion is $$\frac{(-1)^n\sqrt3}{2(2n)!}$$ and the $h^{2n+1}$ term of the Taylor expansion is $$\frac{(-1)^{n+1}}{2(2n+1)!}.$$

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