`f(x)=xsinx` Find the Maclaurin series for the function.

Expert Answers
marizi eNotes educator| Certified Educator

Maclaurin series  is a special case of Taylor series which is centered at a=0. We follow the formula:

`f(x) =sum_(n=0)^oo (f^n(0))/(n!)x^n`

or

`f(x) = f(0) + (f'(0))/(1!)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3 +(f^4(0))/(4!)x^4 +(f^5(0))/(5!)x^5 +...`

To list of `f^n(x)` up to `n=10` , we may apply the Product rule for differentiation: `d/(dx) (u*v) = u'*v +u*v` '.

`f(x) = xsin(x)`

`f'(x) = xcos(x)+sin(x)`

`f''(x) = 2cos(x)-xsin(x)`

`f'''(x) =- xcos(x)-3sin(x)`

`f^4(x) = xsin(x)-4cos(x)`

`f^5(x) = xcos(x)+5sin(x)`

`f^6(x) = 6cos(x)-xsin(x)`

`f^7(x) = -xcos(x)-7sin(x)`

`f^8(x) = xsin(x)-8cos(x)`

`f^9(x) = xcos(x)+9sin(x)`

`f^(10)(x)= 10*cos(x)-x*sin(x)`

Note: `d/(dx)x=1` ,  `d/(dx) cos(x) =-sin(x)` , and `d/(dx) sin(x)=cos(x)` .

Plug-in x =0, we get:

`f(0) = 0*sin(0)`

         `=0*0`

         `=0`

`f'(0) = 0*cos(0)+sin(0)`

         `=0*1+0`

         `=0`

`f''(0) = 2cos(0)-0*sin(0)`

           `=2*1-0*0`

           `=2`

`f'''(0) =- 0*cos(0)-3sin(0)`

            `=-0*1-3*0`

            `=0`

`f^4(0) = 0*sin(0)-4cos(0)`

          `=0*0 -4*1`

          `=-4`

`f^5(0) = 0*cos(0)+5sin(0)`

          `=0*1+5*0`

          `=0`

`f^6(0) = 6cos(0)-0*sin(0)`

          `=6*1-0*0`

          `=6`

`f^7(0) = -0*0cos(0)-7sin(0)`

           `=-0*1-7*0`

           `=0`

`f^8(0) =0*sin(0)-8cos(0)`

          `=0*0-8*1`

          `=-8`

`f^9(0) = 0*cos(0)+9sin(0)`

           `=0*1+9*0`

           `=0`

`f^(10)(0)= 10*cos(0)-0*sin(0)`

            `=10*1-0*0`

            `=10`

Note: `cos(0)=1` and `sin(0) =0` .

Plug-in the values in the formula, we get:

`f(x) = 0 + 0/(1!)x+2/(2!)x^2+0/(3!)x^3+(-4)/(4!)x^4+0/(5!)x^5`

`+6/(6!)x^6+ 0/(7!)x^7+(-8)/(8!)x^8+0/(9!)x^9 +10/(10!)x^10+...`

 

`=0 + 0/(1)x+2/(1*2)x^2+0/(1*2*3)x^3-4/(1*2*3*4)x^4 `

`+ 0/(1*2*3*4*5)x^5 + 6/(1*2*3*4*5*6)x^6+0/(1*2*3*4*5*6*7)x^7`

` -8/(1*2*3*4*5*6*7*8)x^8 + 0/(1*2*3*4*5*6*7*8*9)x^9 + 10/(1*2*3*4*5*6*7*8*9*10)x^(10)+...`

 

`=0 + 0+2/2x^2+0/6x^3-4/24x^4 + 0/120x^5 + 6/720x^6 `

`+0/5040x^7 -8/40320x^8 + 0/362880x^9 +10/3628800x^(10)+...`

 

`=0 + 0+x^2+0-1/6x^4 + 0 + 1/120x^6`

`+0-1/5040x^8 + 0+1/362880x^(10)+...`

 

`=x^2 -1/6x^4 + 1/120x^6 -1/5040x^8 +1/362880x^(10)+...`

Therefore, the Maclaurin series for the function` f(x) =xsin(x)` can be expressed as:

`xsin(x)=x^2 -1/6x^4 + 1/120x^6 -1/5040x^8 +1/362880x^(10)+...`