Question

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

Answer 1

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

Explanation:

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

`\:f\left(x\right)=f\left(a\right)+(f^'\left(a\right))/(1!)\left(x-a\right)+(f^('')\left(a\right))/(2!)\left(x-a\right)^2+(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)+(f^'\left(0\right))/(1!)\left(x\right)+(f^('')\left(0\right))/(2!)\left(x\right)^2+(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(0)/(0!)x^(0)+(0)/(1!)x^(1)+(-4)/(2!)x^(2)+(0)/(3!)x^(3)+(32)/(4!)x^(4)+(0)/(5!)x^(5)+(-192)/(6!)x^(6)+(0)/(7!)x^(7)+(1024)/(8!)x^(8)+(0)/(9!)x^(9)+(-5120)/(10!)x^(10)`

Simplify,

`f\left(x\right)\approx P\left(x\right) = -2x^(2)+(4)/(3)x^(4)- (4)/(15)x^(6)+(8)/(315)x^(8)- (4)/(2835)x^(10)`