Question

How do you use differentiation to find power series representation for $ f\left( x \right) = \dfrac{1}{{{{\left( {1 + x} \right)}^2}}} $ ?

Answer 1

Hint: In the given problem, we are required to find the power series representation for $ f\left( x \right) = \dfrac{1}{{{{\left( {1 + x} \right)}^2}}} $ by d=using differentiation with respect to x. Basics of differentiation should be remembered in order to deal with such problems. Also, rules of differentiation such as power rule, quotient rule, product rule and chain rule must be clear in mind to tackle such conceptual questions.

Complete step-by-step answer:
In the given question, we have to find that the power series representation of $ f\left( x \right) = \dfrac{1}{{{{\left( {1 + x} \right)}^2}}} $ .
We know that $ \dfrac{1}{{\left( {1 + x} \right)}} $ is the resultant expression for the power series: $ 1 - x + {x^2} - .... + {\left( { - x} \right)^k} $ since we know that the sum of infinite geometric series is $ \left( {\dfrac{a}{{1 - r}}} \right) $ where a is the first term of the geometric progression and r is the common ratio.
So, $ 1 - x + {x^2} - .... + {\left( { - x} \right)^k} = \dfrac{1}{{\left( {1 + x} \right)}} $
 $ \Rightarrow \sum\limits_{k = 0}^{\inf } {{{\left( { - x} \right)}^k}} = \dfrac{1}{{\left( {1 + x} \right)}} $
But this is not a power series representation of $ f\left( x \right) = \dfrac{1}{{{{\left( {1 + x} \right)}^2}}} $ . So, we will have to differentiate both sides of the equation with respect to x.
 $ \Rightarrow \dfrac{d}{{dx}}\left[ {\dfrac{1}{{\left( {1 + x} \right)}}} \right] = \dfrac{d}{{dx}}\left[ {1 - x + {x^2} - .... + {{\left( { - x} \right)}^k}} \right] $
So, we differentiate the left side of the equation using the quotient rule of differentiation.
According to quotient rule of differentiation, $ \dfrac{d}{{dx}}\left[ {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \left[ {\dfrac{{g\left( x \right)f'\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{{\left[ {g\left( x \right)} \right]}^2}}}} \right] $ . So, we get,
 $ \Rightarrow \dfrac{{\left( {1 + x} \right)\left( 0 \right) - 1\left( 1 \right)}}{{{{\left( {1 + x} \right)}^2}}} = \dfrac{d}{{dx}}\left[ {1 - x + {x^2} - .... + {{\left( { - x} \right)}^k}} \right] $
Differentiating right side of the equation and simplifying the expression in left side of the equation, we get,
 $ \Rightarrow \dfrac{{ - 1}}{{{{\left( {1 + x} \right)}^2}}} = 0 - 1 + 2x - 3{x^2} + 4{x^3} + ...k{x^{\left( {k - 1} \right)}}{\left( { - 1} \right)^k} $
Simplifying both sides of the equation, we get,
 $ \Rightarrow \dfrac{{ - 1}}{{{{\left( {1 + x} \right)}^2}}} = - 1 + 2x - 3{x^2} + 4{x^3} + ...k{x^{\left( {k - 1} \right)}}{\left( { - 1} \right)^k} $
Multiplying both sides of the equation, we get,
 $ \Rightarrow \dfrac{1}{{{{\left( {1 + x} \right)}^2}}} = 1 - 2x + 3{x^2} - 4{x^3} + ...k{x^{\left( {k - 1} \right)}}{\left( { - 1} \right)^{k + 1}} $
Now, we know that $ {\left( { - 1} \right)^2} = 1 $ . So, we get,
 $ \Rightarrow \dfrac{1}{{{{\left( {1 + x} \right)}^2}}} = 1 - 2x + 3{x^2} - 4{x^3} + ...k{x^{\left( {k - 1} \right)}}{\left( { - 1} \right)^{k - 1}}{\left( { - 1} \right)^2} $
Simplifying the expression further,
 $ \Rightarrow \dfrac{1}{{{{\left( {1 + x} \right)}^2}}} = 1 - 2x + 3{x^2} - 4{x^3} + ...k{\left( { - x} \right)^{\left( {k - 1} \right)}} $
So, we get the power representation of the expression $ f\left( x \right) = \dfrac{1}{{{{\left( {1 + x} \right)}^2}}} $ as $ 1 - 2x + 3{x^2} - 4{x^3} + ...k{\left( { - x} \right)^{\left( {k - 1} \right)}} $ using differentiation.
 $ 1 - 2x + 3{x^2} - 4{x^3} + ...k{\left( { - x} \right)^{\left( {k - 1} \right)}} $ can also be condensed as $ \sum\limits_{k = 0}^{\inf } {k{{\left( { - x} \right)}^{k - 1}}} $
So, the correct answer is “ $ \sum\limits_{k = 0}^{\inf } {k{{\left( { - x} \right)}^{k - 1}}} $ ”.

Note: The given problem is a typical question which involves multiple concepts of mathematics like concepts of series and progressions, binomial theorem, and differentiation. One must be careful while doing the complex and tedious calculations while doing differentiation.