Maclurin expansion

quailmaster SG

A = (n+ 0.5)ln(1 + 1/n) - 1

Use a series expansion to show that if n is large enough for 1/n^3 and higher powers of 1/n to be neglected, A approximately equals to 1/12n^2

Thanks in advance for your help

wee_ws SG

Hi,

Use the fact that ln(1 + x) = x - x^2/2 + x^3/3 - ... from the formulae list.

Thanks!

Cheers,
Wen Shih

quailmaster SG

yeah i tried that and got -1/4n^2 as the ans

(n + 0.5) [ 1/n - 1/2n^2] -1

did i make some careless mistake somewhere?

TrueHeart SG

yeah i tried that and got -1/4n^2 as the ans

(n + 0.5) [ 1/n - 1/2n^2] -1

did i make some careless mistake somewhere?

Hi quailmaster,

When you expand out In(1+(1/n)), you will have to include 1/(3n^3) as well. It is only after simplifying the entire term, then should you get rid of terms including 1/n^3.

Thus, you should get (1/3n^3)(n) - (1/2)(1/2n^2) after the other terms cancel each other out. This will simplify to 1/12n^2.

Cheers.