Reggie

2021-03-11

For $\mathrm{sin}2x$, find the first four terms in the Maclaurin series.
(a) by using the definition of Maclaurin series.
(b) by replacing x by 2x in the series for $\mathrm{sin}2x$.
(c) by multiplying 2 by the series for $\mathrm{sin}x$ by the series for  $\mathrm{cos}x$, because $\mathrm{sin}2x=2$ $\mathrm{sin}x\mathrm{cos}x$

toroztatG

(a)
Calculate the derivative for $f\left(x\right)=\mathrm{sin}\left(2x\right)$ and the the value for the derivatives till 8th order.
$f\left(x\right)=\mathrm{sin}2x$
${f}_{1}\left(x\right)=2\mathrm{cos}2x$
${f}_{2}\left(x\right)=-4\mathrm{sin}2x$
${f}_{3}\left(x\right)=-8\mathrm{cos}2x$
${f}_{4}\left(x\right)=16\mathrm{sin}\left(2x\right)$
${f}_{5}\left(x\right)=32\mathrm{cos}\left(2x\right)$
${f}_{6}\left(x\right)=-64\mathrm{sin}\left(2x\right)$
${f}_{7}\left(x\right)=-128\mathrm{cos}\left(2x\right)$
${f}_{8}\left(x\right)=256\mathrm{sin}\left(2x\right)$
At $x=0$, $f\left(0\right)=\mathrm{sin}\left(2×0\right)=0$
${f}_{1}\left(x\right)=2\mathrm{cos}\left(2×0\right)=2$
${f}_{2}\left(x\right)=-4\mathrm{sin}\left(2×0\right)=0$
${f}_{3}\left(x\right)=-8\mathrm{cos}\left(2×0\right)=-8$
${f}_{4}\left(x\right)=16\mathrm{sin}\left(2×0\right)=0$
${f}_{5}\left(x\right)=32\mathrm{cos}\left(2×0\right)=32$
${f}_{6}\left(x\right)=-64\mathrm{sin}\left(2×0\right)=0$
${f}_{7}\left(x\right)=-128\mathrm{cos}\left(2×0\right)=-128$
${f}_{8}\left(x\right)=256\mathrm{sin}\left(2×0\right)=0$
Now, consider the formula for the maclaurin's series and express f(x) as Maclaurin's series.
Maclaurin's series expansion,
$f\left(x\right)=f\left(0\right)+\frac{{f}_{1}\left(0\right)x}{1!}+\frac{{f}_{2}\left(0\right){x}^{2}}{2!}+\frac{{f}_{3}\left(0\right){x}^{3}}{3!}+\frac{{f}_{4}\left(0\right){x}^{4}}{4!}+\frac{{f}_{5}\left(0\right){x}^{5}}{5!}+\frac{{f}_{6}\left(0\right){x}^{6}}{6!}+\frac{{f}_{7}\left(0\right){x}^{7}}{7!}+\frac{{f}_{8}\left(0\right){x}^{8}}{8!}+\frac{{f}_{9}\left(0\right){x}^{9}}{9!}$
$=0+\frac{{f}_{1}\left(0\right)x}{1!}+0+\frac{{f}_{3}\left(0\right){x}^{3}}{3!}+0+\frac{{f}_{5}\left(0\right){x}^{5}}{5!}+0+\frac{{f}_{7}\left(0\right){x}^{7}}{7!}+0$
$=\frac{2x}{1!}+\frac{\left(-8\right){x}^{3}}{3!}+\frac{\left(32\right){x}^{5}}{5!}+\frac{\left(-128\right){x}^{7}}{7!}$
So, $f\left(x\right)=2x-\frac{4{x}^{3}}{3}+\frac{4}{15}{x}^{5}-\frac{8}{315}{x}^{7}$
Hence, $\mathrm{sin}2x=2x-\frac{4{x}^{3}}{3}+\frac{4}{15}{x}^{5}-\frac{8}{315}{x}^{7}$
(b) Now, replace x by 2x and evaluate the series for $\mathrm{sin}4x$.

Jeffrey Jordon