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So I was given the following prompt:

"For the following, use the definition to find the Taylor (or Maclaurin) series centered at c for the function. When writing your answers, be sure to list the first 4 non-zero terms and the general term."

$$f(x)=\sin(2x), \ \ \ c=0.$$

I guess I'm confused on how I'd find the general term from this, I worked out the Maclaurin series to look something like the following: $P_n(x)=2x-\frac{4}{3}x^3+\frac{4}{15}x^5-\frac{8}{315}x^7$. I understand that I needed to look for the pattern here to find the general term, but I don't know where I'd start for that and I also didn't know if there was an easier way to do this through a formula. Any help would be appreciated!

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  • $\begingroup$ If it is hard to guess from the explicit coefficients, use instead the formula for the coefficients in terms of the successive derivatives of $f$ evaluated at $0$. The coefficients have the form $f^{(n)}(0)/n!$. Compute a few of the derivatives (before evaluating at $x=0$) and you will see its form. $\endgroup$
    – plop
    May 11 '21 at 14:23
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Your Maclauren series is correct. If you look at how you got the terms before you canceled the $2$s to reduce to lowest terms you can see the general term is $\pm \frac {2^n}{n!}$.
That doesn't get you a Taylor series centered at $c$. Clearly the constant term is $\sin 2c$ and you will have terms of all orders because the sine function is not odd around $c$ in general. You can either use the Taylor series formula or write $$\sin(2(c+x))=\sin(2c)\cos (2x) + \cos(2c) \sin (2x)$$
and use the Maclauren series for both sine and cosine modified as above for the factor $2$

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When you have the problem of $\lim_{x\to c} \, f(x)$ or Taylor expansions around $x=c$, it makes life easier to let $x=y-c$ in order to work around $y=0$.

Doing what @Ross Millikan answered, in the most compact form, you should get $$\sin(2x)=\sum_{n=0 }^\infty \frac{2^n \sin \left(2 c+n\frac{\pi }{2}\right)}{n!} (x-c)^n$$

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You can make computing the Taylor series somewhat easier by considering the complex exponential,

$$e^{inx}=\cos(nx)+i\sin(nx)$$

where $i=\sqrt{-1}$. Let $n=2$; then the imaginary parts of $e^{2ix}$ and its derivatives make up the Taylor expansion of $f(x)$.

Written in this way, the derivatives are a bit more convenient to compute:

$$\frac{\mathrm de^{2ix}}{\mathrm dx}=2ie^{2ix},\frac{\mathrm d^2e^{2ix}}{\mathrm dx}=(2i)^2e^{2ix},\frac{\mathrm d^3e^{2ix}}{\mathrm dx}=(2i)^3e^{2ix},\cdots$$

and the rule for the $n$th term for the expansion around $x=c$ is clearly $\frac{(2i)^ne^{2ic}}{n!}x^n$, which in turn means the $n$th term in the Taylor series of $f(x)$ is $\operatorname{Im}\left(\frac{(2i)^ne^{2ic}}{n!}\right)$.

At $c=0$, $e^{2ic}=1$. For even $n$, you have $i^n\in\{-1,1\}$; for odd $n$, $i^n\in\{-i,i\}$. So you only care about odd-indexed terms in the expansion of $e^{2ix}$. Let $n=2k-1$ for integers $k\ge1$:

$$\operatorname{Im}\left(\frac{(2i)^ne^{2ic}}{n!}\right)=\operatorname{Im}\left(\frac{2^{2k-1}i^{2k-1}}{(2k-1)!}\right)=\operatorname{Im}\left(-i\frac{2^{2k-1}i^{2k}}{(2k-1)!}\right)=\boxed{\frac{(-1)^{k+1}2^{2k-1}}{(2k-1)!}}$$

Taking $k\in\{1,2,3,4\}$, you get the same first four non-zero terms of the expansion:

$$\begin{align} f(x) &\approx \frac{(-1)^{1+1} 2^1}{1!}x + \frac{(-1)^{2+1} 2^3}{3!}x^3 + \frac{(-1)^{3+1} 2^5}{5!}x^5 + \frac{(-1)^{4+1} 2^7}{7!}x^7\\[1ex] &=2x-\frac{4x^3}3+\frac{4x^5}{15}-\frac{8x^7}{315} \end{align}$$

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