Question

Expand $ {{\left( 1-2x-2{{x}^{2}} \right)}^{\dfrac{1}{4}}} $ as far as $ {{x}^{2}} $ .\[\]

Answer 1

Hint: We recall binomial theorem for integral index then we recall the binomial series for fractional and negative indices $ {{\left( 1+x \right)}^{\alpha }}=\sum\limits_{k=0}^{\infty }{{}^{\alpha }{{C}_{k}}{{x}^{k}}}=1+\alpha x+\dfrac{\alpha \left( \alpha -1 \right)}{2!}{{x}^{2}}+... $ . We use binomial series for fractional index from the question $ \alpha =\dfrac{1}{4} $ until $ {{x}^{2}} $ terms appear in the expansion.

Complete step by step answer:
We know that binomial is the algebraic expression involving two terms and each term with distinct mathematical objects. We know that we can the binomial theorem (or binomial expansion) describes the algebraic expansion of power of binomial. If $ x,y $ are the two terms of binomial with some positive integral power $ n $ then the binomial expansion is given by;
\[{{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{y}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{y}^{0}}+{}^{n}{{C}_{2}}{{x}^{n-2}}{{y}^{0}}+...+{}^{n}{{C}_{n}}{{x}^{0}}{{y}^{n}}\]
The above expression is called binomial formula or binomial identity. The combinatorial terms $ {}^{n}{{C}_{0}},{}^{n}{{C}_{1}},{}^{n}{{C}_{2}},...,{}^{n}{{C}_{n}} $ are called binomial coefficients. The above binomial expansion is finite and valid for only integral indices. We use binomial series for any index $ \alpha $ as expresses as a power series as
\[{{\left( 1+x \right)}^{\alpha }}=\sum\limits_{k=0}^{\infty }{{}^{\alpha }{{C}_{k}}{{x}^{k}}}=1+\alpha x+\dfrac{\alpha \left( \alpha -1 \right)}{2!}{{x}^{2}}+...\]
 The general binomial coefficient is given by ;
\[{}^{\alpha }{{C}_{k}}=\dfrac{\alpha \left( \alpha -1 \right)\left( \alpha -2 \right)...\left( \alpha -k+1 \right)}{k!}\]

We are asked in the question to expand $ {{\left( 1-2x-2{{x}^{2}} \right)}^{\dfrac{1}{4}}} $ as far as $ {{x}^{2}} $ . It means we have to find the sum of terms in the expansion as long as $ {{x}^{2}} $ terms appear in the expansion of $ {{\left( 1-2x-2{{x}^{2}} \right)}^{\dfrac{1}{4}}} $ . So let us consider
\[{{\left( 1-2x-2{{x}^{2}} \right)}^{\dfrac{1}{4}}}={{\left( 1-\left( 2x+2{{x}^{2}} \right) \right)}^{\dfrac{1}{4}}}\]
We expand taking $ \alpha =\dfrac{1}{4} $ and $ x=2x+2{{x}^{2}} $ . We have the first term in the binomial expansion as 1. We have second term as
\[\begin{align}
  & \alpha \times \left( -\left( 2x+2{{x}^{2}} \right) \right)=-\dfrac{1}{4}\left( 2x+2{{x}^{2}} \right) \\
 & \Rightarrow \alpha \times \left( -\left( 2x+2{{x}^{2}} \right) \right)=-\dfrac{1}{2}x-\dfrac{1}{2}{{x}^{2}} \\
\end{align}\]
We have the third term as
\[\begin{align}
  & \dfrac{\alpha \left( \alpha -1 \right)}{2!}{{\left( 2x+2{{x}^{2}} \right)}^{2}}=\dfrac{\dfrac{1}{4}\left( 1-\dfrac{1}{4} \right)}{2}\left( 4{{x}^{2}}+4{{x}^{4}}+8{{x}^{3}} \right) \\
 & \Rightarrow \dfrac{\alpha \left( \alpha -1 \right)}{2!}{{\left( 2x+2{{x}^{2}} \right)}^{2}}=\dfrac{\dfrac{1}{4}\times \dfrac{-3}{4}}{2}\left( 4{{x}^{2}}+4{{x}^{4}}+8{{x}^{3}} \right) \\
 & \Rightarrow \dfrac{\alpha \left( \alpha -1 \right)}{2!}{{\left( 2x+2{{x}^{2}} \right)}^{2}}=\dfrac{-3}{32}\left( 4{{x}^{2}}+4{{x}^{4}}+8{{x}^{3}} \right) \\
 & \Rightarrow \dfrac{\alpha \left( \alpha -1 \right)}{2!}{{\left( 2x+2{{x}^{2}} \right)}^{2}}=-\dfrac{3}{8}{{x}^{2}}-\dfrac{3}{8}{{x}^{4}}-\dfrac{3}{4}{{x}^{3}} \\
\end{align}\]
We have the fourth term as;
\[\dfrac{\alpha \left( \alpha -1 \right)\left( \alpha -2 \right)}{3!}{{\left( 2x+2{{x}^{2}} \right)}^{3}}\]
We see that from fourth term onwards the exponents on $ x $ will always be greater than 2. So we shall not get any term with $ {{x}^{2}} $ after the third term. We collect terms as far as $ {{x}^{2}} $ from the first three terms to have;
\[{{\left( 1-2x-2{{x}^{2}} \right)}^{\dfrac{1}{4}}}=1-\dfrac{1}{2}x-\dfrac{1}{2}{{x}^{2}}-\dfrac{3}{8}{{x}^{2}}=1-\dfrac{1}{2}x-\dfrac{7}{8}{{x}^{2}}\]


Note:
We note that the binomial series converges only if $ \left| x \right|\le 1 $ and $ \left| \alpha \right| < 1 $ . We can also apply the binomial series for complex numbers. If the question would have asked to expand as far as $ {{x}^{3}} $ we would have expanded up to the fourth term. We can take the limit to find the exponential series and then the logarithmic series from the binomial series.