Question

Expand the function using Maclaurin Series Expansion (first 8 terms).

1. (1+x)^1/2 , x=5

Answer 1

Solution

According to definition of Maclaurin Series Expansion

$f(x)=\sum_{n=0}^\infty f^{(n)}(0) x^n / n!$

For f(x)=(1+x)^1/2   f’(x)=(1/2) (1+x)^-1/2 ,  f’’(x)=-(1/2)² (1+x)^-3/2 ,  f’’’(x)=(1/2)³ 3(1+x)^-5/2 , f⁽⁴⁾(x)=-(1/2)⁴ 3*5*(1+x)^-7/2 , …, f⁽ⁿ⁾(x)=(-)^n-1(1/2)ⁿ 3*5*…*(2n-3)*(1+x)^-(2n-1)/2     

So

$(1+x)^{1/2}=1+x/2+ \sum_{n=2}^\infty (-)^{n-1}(2n-3)!! x^n / 2n!!$

For first 8 terms

(1+x)^1/2 ≈ 1+x/2-x²/8+x³/16-5x⁴/128+7x⁵/256-21x⁶/1024+33x⁷/2048

For x=5

(1+x)^1/2 = 2.449,

1+x/2-x²/8+x³/16-5x⁴/128+7x⁵/256-21x⁶/1024+33x⁷/2048 = 1088