How do you find the Maclaurin Series for #sinx-x #?

Question

2225 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-24T10:46:58

#sinx - x = sum_(n=1)^oo (-1)^n x^(2n+1)/((2n+1)!)#

Determine the MacLaurin series for #sin x#:

#d/dx sinx = cosx = sin(x+pi/2)#

#d^2/dx^2 sinx = d/dx cosx = d/dx sin(x+pi/2) = cos(x+pi/2) = sin(x+pi)#

and in general:

#d^n/dx^n = sin(x+(npi)/2)#

so that the coefficient are zero for #n# even, and for #n# odd:

# [d^(2n+1)/dx^(2n+1) sinx]_(x=0) = (-1)^n#

Then:

#sinx = sum_(n=0)^oo (-1)^n x^(2n+1)/((2n+1)!)#

Isolate the term for #n=0#:

#sinx = x +sum_(n=1)^oo (-1)^n x^(2n+1)/((2n+1)!)#

and finally:

#sinx - x = sum_(n=1)^oo (-1)^n x^(2n+1)/((2n+1)!)#