Use the binomial series to expand the function as a power series.4√(1 - x).

Solution:

Given, the function is f(x) = \((1-x)^\frac{1}{4}\)

We have to use binomial series to expand the function as a power series.

The general form for raising a binomial to a whole number power is given by

\((a+b)^{n}=\sum_{k=0}^{n}(\frac{n}{k})a^{(n-k)}b^{k}\)

Where, \(\frac{n}{k}=\frac{n!}{k!(n-k)!)}=\frac{n(n-1).....(n-k+1)}{k!}\)

So, \((1+x)^{n}=1+nx+\frac{n(n-1)x^{2}}{2!}+\frac{n(n-1)(n-2)x^{3}}{3!}+.............\)

Here, n = 1/3, x = -x

\(\\(1-x)^{\frac{1}{4}}=1+(\frac{1}{4})(-x)+\frac{\frac{1}{4}(\frac{1}{4}-1)(-x)^{2}}{2\times 1}+\frac{\frac{1}{4}(\frac{1}{4}-1)(\frac{1}{4}-2)(-x)^{3}}{3\times 2\times 1}+.............\\=1+\frac{1}{4}(-x)+\frac{\frac{1}{4}(\frac{-3}{4})x^{2}}{2}+\frac{\frac{1}{4}(\frac{-3}{4})(\frac{-7}{4})(-x)^{3}}{6}+....\\=1-(\frac{1}{4})x-(\frac{3}{32})x^{2}-(\frac{7}{128})x^{3}+........\)

Therefore, \(\\(1-x)^{\frac{1}{4}}=1-(\frac{1}{4})x-(\frac{3}{32})x^{2}-(\frac{7}{128})x^{3}+........\).

Use the binomial series to expand the function as a power series.4√(1 - x).

Summary:

Using the binomial series to expand the function as a power series.4√(1 - x) is \(1-(\frac{1}{4})x-(\frac{3}{32})x^{2}-(\frac{7}{128})x^{3}+........\).