Question

How do you use the Maclaurin series for `f(x) = (2+x)^5`?

Answer 1

`f(x)=32+80x+80x^2+40x^3+10x^4+x^5`

Explanation:

According to Maclaurin series,

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

As `f(x)=(2+x)^5`, `f(0)=32`

`f'(x)=5(2+x)^4` and `f''(0)=80`

`f''(x)=5xx4(2+x)^3` and `f'''(0)=160`

`f'''(x)=5xx4xx3(2+x)^2` and `f''''(0)=240`

`f''''(x)=5xx4xx3xx2(2+x)` and `f''''(0)=240`

`f'''''(x)=5xx4xx3xx2xx1` and `f''''(0)=120` and `f''''''(x)=0`

Hence, `f(x)=32+80x+160x^2/2+240x^3/6+240x^4/24+120x^5/120`

or `f(x)=32+80x+80x^2+40x^3+10x^4+x^5`