Question

Use binomial series to expand $f\left( x \right) = \sqrt {1 + {x^2}} $?

Answer 1

Hint: In this question we have to expand the given expression by using binomial formula, as the given power has a fractional exponent we will use the binomial series for fractions which is given by,
${\left( {1 + x} \right)^n} = 1 + \dfrac{{nx}}{{1!}} + \dfrac{{n\left( {n - 1} \right){x^2}}}{{2!}} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right){x^3}}}{{3!}} + ....$ , Now substituting the values from the given expression in the question and we will get the required result.

Complete step by step answer:
Binomial expansion for the fractional exponent is given by the formula, ${\left( {1 + x} \right)^n} = 1 + \dfrac{{nx}}{{1!}} + \dfrac{{n\left( {n - 1} \right){x^2}}}{{2!}} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right){x^3}}}{{3!}} + ....$,
So, now given expression is$f\left( x \right) = \sqrt {1 + {x^2}} $,
Rewriting the given expression we get,
$ \Rightarrow f\left( x \right) = {\left( {1 + {x^2}} \right)^{\dfrac{1}{2}}}$,
Now substituting the values in the expansion, here $1 = 1$, $x = {x^2}$and $n = \dfrac{1}{2}$, then we get,
$ \Rightarrow {\left( {1 + {x^2}} \right)^{\dfrac{1}{2}}} = 1 + \dfrac{{\left( {\dfrac{1}{2}} \right){x^2}}}{{1!}} + \dfrac{{\dfrac{1}{2}\left( {\dfrac{1}{2} - 1} \right){{\left( {{x^2}} \right)}^2}}}{{2!}} + \dfrac{{\dfrac{1}{2}\left( {\dfrac{1}{2} - 1} \right)\left( {\dfrac{1}{2} - 2} \right){{\left( {{x^2}} \right)}^3}}}{{3!}} + .....$,
Now applying factorial formula i.e.,$n! = n \times (n - 1) \times \left( {n - 2} \right) \times ........ \times 3 \times 2 \times 1$,we get,
$ \Rightarrow {\left( {1 + {x^2}} \right)^{\dfrac{1}{2}}} = 1 + \dfrac{{\left( {\dfrac{1}{2}} \right){x^2}}}{1} + \dfrac{{\dfrac{1}{2}\left( { - \dfrac{1}{2}} \right){x^4}}}{{1 \times 2}} + \dfrac{{\dfrac{1}{2}\left( { - \dfrac{1}{2}} \right)\left( { - \dfrac{3}{2}} \right){x^6}}}{{1 \times 2 \times 3}} + .....$
Now simplifying the expansion by multiplying the fractions in the numerator, we get,
$ \Rightarrow {\left( {1 + {x^2}} \right)^{\dfrac{1}{2}}} = 1 + \dfrac{1}{2}{x^2} + \dfrac{{ - \dfrac{1}{4}{x^4}}}{{1 \times 2}} + \dfrac{{\dfrac{1}{2}\left( {\dfrac{3}{4}} \right){x^6}}}{{1 \times 2 \times 3}}.....$
Now simplifying by multiplying and dividing we get,
$ \Rightarrow {\left( {1 + {x^2}} \right)^{\dfrac{1}{2}}} = 1 + \dfrac{1}{2}{x^2} - \dfrac{{1{x^4}}}{{1 \times 2 \times 4}} + \dfrac{{1{x^6}}}{{1 \times 2 \times 2 \times 4}}.....$
Now final simplification we get,
$ \Rightarrow {\left( {1 + {x^2}} \right)^{\dfrac{1}{2}}} = 1 + \dfrac{1}{2}{x^2} - \dfrac{{{x^4}}}{8} + \dfrac{{{x^6}}}{{16}}.....$.
So, the expansion of the given expression will be $1 + \dfrac{1}{2}{x^2} - \dfrac{{{x^4}}}{8} + \dfrac{{{x^6}}}{{16}}.....$.

$\therefore $ The expansion form of the given expression $f\left( x \right) = \sqrt {1 + {x^2}} $ using binomial series will be equal to$1 + \dfrac{1}{2}{x^2} - \dfrac{{{x^4}}}{8} + \dfrac{{{x^6}}}{{16}}.....$

Note: We use the binomial theorem to help us expand binomials to any given power without direct multiplication. As we have seen, multiplication can be time-consuming or even not possible in some cases.
The binomial theorem is an algebraic method of expanding a binomial expression. Essentially, it demonstrates what happens when you multiply a binomial by itself. For example, consider the expression ${\left( {2x + 5} \right)^9}$. It would take quite a long time to multiply the binomial $2x + 5$ out nine times. The binomial theorem provides a short cut, or a formula that yields the expanded form of this expression.