Answer to Question #156959 in Civil and Environmental Engineering for Ian

Question #156959

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

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


1
Expert's answer
2021-01-26T02:58:27-0500

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+\n\\sum_{n=2}^\\infty \n(-)^{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


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