Question

Write the first three nonzero terms in the Maclaurin seriesfor xsin(-2x).

Answer 1

Answer:

The Maclaurin of $xsin\left(-2x\right)$ is $-2x^2+\frac{4}{3}x^4-\frac{4}{15}x^6+\frac{8}{315}x^8-\frac{4}{2835}x^{10}+\ldots$.

Step-by-step explanation:

Taylor series of function $f\left(x\right)$ at a is defined as:

$\:f\left(x\right)=f\left(a\right)+\frac{f^'\left(a\right)}{1!}\left(x-a\right)+\frac{f^{''}\left(a\right)}{2!}\left(x-a\right)^2+\frac{f^{'''}\left(a\right)}{3!}\left(x-a\right)^3+\ldots$

Maclaurin series of function $f\left(x\right)$ is a Taylor series  of function $f\left(x\right)$ at a = 0

$\:f\left(x\right)=f\left(0\right)+\frac{f^'\left(0\right)}{1!}\left(x\right)+\frac{f^{''}\left(0\right)}{2!}\left(x\right)^2+\frac{f^{'''}\left(0\right)}{3!}\left(x\right)^3+\ldots$

Step 1: Find the derivatives of $f\left(x\right)=x\sin \left(-2x\right)$ at a = 0

$f^{(1)}\left(x\right)=\left(f^{(0)}\left(x\right)\right)^{\prime}=\left(- x \sin{\left(2 x \right)}\right)^{\prime}=- 2 x \cos{\left(2 x \right)} - \sin{\left(2 x \right)}$

$f^{(2)}\left(x\right)=\left(f^{(1)}\left(x\right)\right)^{\prime}=\left(- 2 x \cos{\left(2 x \right)} - \sin{\left(2 x \right)}\right)^{\prime}=4 x \sin{\left(2 x \right)} - 4 \cos{\left(2 x \right)}$

$f^{(3)}\left(x\right)=\left(f^{(2)}\left(x\right)\right)^{\prime}=\left(4 x \sin{\left(2 x \right)} - 4 \cos{\left(2 x \right)}\right)^{\prime}=8 x \cos{\left(2 x \right)} + 12 \sin{\left(2 x \right)}$

$f^{(4)}\left(x\right)=\left(f^{(3)}\left(x\right)\right)^{\prime}=\left(8 x \cos{\left(2 x \right)} + 12 \sin{\left(2 x \right)}\right)^{\prime}=- 16 x \sin{\left(2 x \right)} + 32 \cos{\left(2 x \right)}$

$f^{(5)}\left(x\right)=\left(f^{(4)}\left(x\right)\right)^{\prime}=\left(- 16 x \sin{\left(2 x \right)} + 32 \cos{\left(2 x \right)}\right)^{\prime}=- 32 x \cos{\left(2 x \right)} - 80 \sin{\left(2 x \right)}$

$f^{(6)}\left(x\right)=\left(f^{(5)}\left(x\right)\right)^{\prime}=\left(- 32 x \cos{\left(2 x \right)} - 80 \sin{\left(2 x \right)}\right)^{\prime}=64 x \sin{\left(2 x \right)} - 192 \cos{\left(2 x \right)}$

$f^{(7)}\left(x\right)=\left(f^{(6)}\left(x\right)\right)^{\prime}=\left(64 x \sin{\left(2 x \right)} - 192 \cos{\left(2 x \right)}\right)^{\prime}=128 x \cos{\left(2 x \right)} + 448 \sin{\left(2 x \right)}$

$f^{(8)}\left(x\right)=\left(f^{(7)}\left(x\right)\right)^{\prime}=\left(128 x \cos{\left(2 x \right)} + 448 \sin{\left(2 x \right)}\right)^{\prime}=- 256 x \sin{\left(2 x \right)} + 1024 \cos{\left(2 x \right)}$

$f^{(9)}\left(x\right)=\left(f^{(8)}\left(x\right)\right)^{\prime}=\left(- 256 x \sin{\left(2 x \right)} + 1024 \cos{\left(2 x \right)}\right)^{\prime}=- 512 x \cos{\left(2 x \right)} - 2304 \sin{\left(2 x \right)}$

$f^{(10)}\left(x\right)=\left(f^{(9)}\left(x\right)\right)^{\prime}=\left(- 512 x \cos{\left(2 x \right)} - 2304 \sin{\left(2 x \right)}\right)^{\prime}=1024 x \sin{\left(2 x \right)} - 5120 \cos{\left(2 x \right)}$

Step 2: Evaluate the derivatives at the given point.

$\left(f\left(0\right)\right)^{\prime }=0$

$\left(f\left(0\right)\right)^{\prime \prime }=-4$

$\left(f\left(0\right)\right)^{\prime \prime \prime }=0$

$\left(f\left(0\right)\right)^{\prime \prime \prime \prime }=32$

$\left(f\left(0\right)\right)^{\left(5\right)}=0$

$\left(f\left(0\right)\right)^{\left(6\right)}=-192$

$\left(f\left(0\right)\right)^{\left(7\right)}=0$

$\left(f\left(0\right)\right)^{\left(8\right)}=1024$

$\left(f\left(0\right)\right)^{\left(9\right)}=0$

$\left(f\left(0\right)\right)^{\left(10\right)}=-5120$

Step 3: Use the calculated values to get a polynomial

$f\left(x\right)\approx\frac{0}{0!}x^{0}+\frac{0}{1!}x^{1}+\frac{-4}{2!}x^{2}+\frac{0}{3!}x^{3}+\frac{32}{4!}x^{4}+\frac{0}{5!}x^{5}+\frac{-192}{6!}x^{6}+\frac{0}{7!}x^{7}+\frac{1024}{8!}x^{8}+\frac{0}{9!}x^{9}+\frac{-5120}{10!}x^{10}$

Simplify,

$f\left(x\right)\approx P\left(x\right) = -2x^{2}+\frac{4}{3}x^{4}- \frac{4}{15}x^{6}+\frac{8}{315}x^{8}- \frac{4}{2835}x^{10}$