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Find the first four nonzero terms of the series for $f(x)$ centered at $a$, using the definition of Taylor series. $$f(x) = \sin(x),\quad a=\pi/6$$

I got this:

1st term: $1/2$

2nd: $\sqrt{3}/2$

3rd: $-1/2$

4th: $-\sqrt{3}/2$

but it seems I am very wrong, when I checked the answer. What am I doing wrong?

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    $\begingroup$ What is the definition of a Taylor series? $\endgroup$ – user137731 Jul 19 '16 at 22:59
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    $\begingroup$ What you call the third term, for example, should be $\frac{1}{2!}(-1/2)(x-\pi/6)^2$. $\endgroup$ – André Nicolas Jul 19 '16 at 23:01
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One may recall that, for any sufficiently regular function $f$, by the Taylor series expansion near $x=a$, one has $$ f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+(x-a)^3\varepsilon(x-a) $$ with $\displaystyle \lim_{x \to a}\varepsilon(x-a)=0$. Here we take $f(x)=\sin x$, $a=\dfrac{\pi}6$, then classically $$ f\left(\frac{\pi}6\right)=\frac12,\quad f'\left(\frac{\pi}6\right)=\frac{\sqrt{3}}2, \quad f''\left(\frac{\pi}6\right)=-\frac12, \quad f'''\left(\frac{\pi}6\right)=-\frac{\sqrt{3}}2, $$ giving

$$ \sin x=\frac12+\frac{\sqrt{3}}2\left(x-\frac{\pi }{6}\right)-\frac14 \left(x-\frac{\pi }{6}\right)^2-\frac{\sqrt{3}}{12}\left(x-\frac{\pi }{6}\right)^3+\left(x-\frac{\pi }{6}\right)^3\varepsilon\left(x-\frac{\pi }{6}\right). $$

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  • $\begingroup$ What if it was a maclaurin series? My a would be 0, how would the rest of the problem change? I would do f(0) instead of f(pi/6), would I have just x, x^2, x^3 instead of x - pi/6, x - pi/6 squared, and so on? $\endgroup$ – dmscs Jul 19 '16 at 23:32
  • $\begingroup$ @dmscs Yes, exactly. $\endgroup$ – Olivier Oloa Jul 19 '16 at 23:34
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Another way to do it could have been to set $x=y+\frac \pi 6$ and use

$$\sin(y+\frac \pi 6)=\frac{\sqrt{3}}{2} \sin (y)+\frac{1}{2}\cos (y)$$ and, now, use Taylor series around $y=0$ $$\sin(y)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}y^{2n+1}$$ $$\cos(y)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}y^{2n}$$ which make $$\sin(y+\frac \pi 6)=\frac{\sqrt{3}}{2}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}y^{2n+1}+\frac{1}{2}\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}y^{2n}$$ Back to $x$

$$\sin(x)=\frac{\sqrt{3}}{2}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(x-\frac \pi 6)^{2n+1}+\frac{1}{2}\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}(x-\frac \pi 6)^{2n}$$

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