## Calculus: Early Transcendentals 8th Edition

(a) $(1+x)^{-1/4}=1+\Sigma_{n=1}^{\infty}(-1)^{n}\frac{1.5.9.13......(4n-3)x^{n}}{4^{n}n!}$ (b) $\frac{1}{\sqrt[4] {1.1}}\approx 1+(-0.025)+(0.0015625)\approx 0.976$
(a) $\frac{1}{\sqrt[4] {1+x}}=(1+x)^{-1/4}$ $(1+x)^{-1/4}=\Sigma_{n=0}^{\infty}(-1/4)nx^{n}$ $\frac{1}{\sqrt[4] {1+x}}=(1+x)^{-1/4}=1+\Sigma_{n=1}^{\infty}(-1)^{n}\frac{1.5.9.13......(4n-3)x^{n}}{4^{n}n!}$ (b) From part (a), we have $\frac{1}{\sqrt[4] {1+x}}=(1+x)^{-1/4}=1+\Sigma_{n=1}^{\infty}(-1)^{n}\frac{1.5.9.13......(4n-3)x^{n}}{4^{n}n!}$ Here $a_{n}=1+\Sigma_{n=1}^{\infty}(-1)^{n}\frac{1.5.9.13......(4n-3)(0.1)^{n}}{4^{n}n!}$ Thus, $\frac{1}{\sqrt[4] {1.1}}\approx 1+(-0.025)+(0.0015625)\approx 0.976$