Question

How do you find the Maclaurin series and radius of convergence for $f\left( x \right) = \dfrac{1}{{{{\left( {1 + x} \right)}^2}}}$?

Answer 1

This question can be solved by recalling the fact that the Maclaurin series is defined by $f\left( 0 \right) + \dfrac{{f'\left( 0 \right)x}}{{1!}} + \dfrac{{f''\left( 0 \right){x^2}}}{{2!}} + \dfrac{{f'''\left( 0 \right){x^3}}}{{3!}} + ........\dfrac{{{f^n}\left( x \right){x^n}}}{{n!}}$, we will find the general form of all derivatives at the function $f\left( x \right)$ and substitute $x = 0$ in the formula and make necessary calculations to get the required result, and the radius of the convergence is given by ratio test i.e., $\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right|$.

Complete step-by-step answer:
We know that the Maclaurin series is given by,
$f\left( 0 \right) + \dfrac{{f'\left( 0 \right)x}}{{1!}} + \dfrac{{f''\left( 0 \right){x^2}}}{{2!}} + \dfrac{{f'''\left( 0 \right){x^3}}}{{3!}} + ........\dfrac{{{f^n}\left( x \right){x^n}}}{{n!}}$
Now the given function is $f\left( x \right) = \dfrac{1}{{{{\left( {1 + x} \right)}^2}}}$,
Now substitute zero in $x$ place we get,
$\Rightarrow f\left( 0 \right) = \dfrac{1}{{{{\left( {1 + 0} \right)}^2}}}$,
Now simplifying we get,
$\Rightarrow f\left( 0 \right) = \dfrac{1}{{{{\left( 1 \right)}^2}}}$,
Now finally we get,
$\Rightarrow f\left( 0 \right) = 1$,
Now deriving the given function using the identity$f\left( x \right) = {x^{ - n}}$then$f'\left( x \right) = - n{x^{ - n - 1}}$ we get,
$f\left( x \right) = \dfrac{1}{{{{\left( {1 + x} \right)}^2}}}$,
$\Rightarrow f'\left( x \right) = \dfrac{{ - 2}}{{{{\left( {1 + x} \right)}^3}}}$,
Now substitute the value of zero in $x$ place we get,
$\Rightarrow f'\left( 0 \right) = \dfrac{{ - 2}}{{{{\left( {1 + 0} \right)}^3}}}$,
Now simplifying we get,
$\Rightarrow f'\left( 0 \right) = \dfrac{{ - 2}}{1} = - 2$,
Now again differentiating $f'\left( x \right) = \dfrac{{ - 2}}{{{{\left( {1 + x} \right)}^3}}}$ we get,
$\Rightarrow f''\left( x \right) = \dfrac{d}{{dx}}\dfrac{{ - 2}}{{{{\left( {1 + x} \right)}^3}}}$,
Now again applying the identity we get,
$\Rightarrow f''\left( x \right) = - 3\left( {\dfrac{{ - 2{{\left( {1 + x} \right)}^2}}}{{{{\left( {1 + x} \right)}^6}}}} \right)$,
Now simplifying we get,
$\Rightarrow f''\left( x \right) = \dfrac{6}{{{{\left( {1 + x} \right)}^4}}}$,
Now substitute the value of zero in $x$ place we get,
$\Rightarrow f''\left( 0 \right) = \dfrac{6}{{{{\left( {1 + 0} \right)}^4}}} = 6$,
Now again differentiating $f''\left( x \right) = \dfrac{6}{{{{\left( {1 + x} \right)}^4}}}$ we get,
$\Rightarrow f'''\left( x \right) = \dfrac{d}{{dx}}\dfrac{6}{{{{\left( {1 + x} \right)}^4}}}$,
Now applying derivative identity, we get,
$\Rightarrow f'''\left( x \right) = 4\left( {\dfrac{{ - 6{{\left( {1 + x} \right)}^3}}}{{{{\left( {1 + x} \right)}^8}}}} \right)$,
Now simplifying we get,
$\Rightarrow f'''\left( x \right) = \dfrac{{ - 24}}{{{{\left( {1 + x} \right)}^5}}}$,
Now substitute the value of zero in $x$ place we get,
$\Rightarrow f'''\left( 0 \right) = \dfrac{{ - 24}}{{{{\left( {1 + 0} \right)}^5}}} = - 24$,
Now substituting these values in the Maclaurin series we get,
$f\left( 0 \right) + \dfrac{{f'\left( 0 \right)x}}{{1!}} + \dfrac{{f''\left( 0 \right){x^2}}}{{2!}} + \dfrac{{f'''\left( 0 \right){x^3}}}{{3!}} + ........\dfrac{{f'\left( 0 \right){x^n}}}{{n!}}$,
$\Rightarrow 1 + \dfrac{{ - 2x}}{{1!}} + \dfrac{{6{x^2}}}{{2!}} - \dfrac{{24{x^3}}}{{3!}} + .....$
Now simplifying we get,
$\Rightarrow 1 - 2x + \dfrac{{6{x^2}}}{2} - \dfrac{{24{x^3}}}{6} + .....$,
Now again simplifying we get,
$\Rightarrow 1 - 2x + 3{x^2} - 4{x^3} + .....$
Which can be written in summation notation as,
$\Rightarrow \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n + 1}}\left( n \right){x^{n - 1}}}$,
Now the radius of the convergence is given by ratio test i.e., $\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right|$,
Now here${a_{n + 1}}$can be found by substituting n by n+1 we get,
$\Rightarrow {a_{n + 1}} = {\left( { - 1} \right)^{n + 1 + 1}}\left( {n + 1} \right){x^{n + 1 - 1}}$,
Now substituting the values in the ratio test formula we get,
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{{{{\left( { - 1} \right)}^{n + 1 + 1}}\left( {n + 1} \right){x^{n + 1 - 1}}}}{{{{\left( { - 1} \right)}^{n + 1}}\left( n \right){x^{n - 1}}}}$,
Now simplifying we get,
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{{{{\left( { - 1} \right)}^{n + 2}}\left( {n + 1} \right){x^n}}}{{{{\left( { - 1} \right)}^{n + 1}}\left( n \right){x^{n - 1}}}}$,
Now again simplifying by using the exponents identities we get,
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{{{{\left( { - 1} \right)}^{n + 2 - n - 1}}\left( {n + 1} \right){x^{n - n + 1}}}}{n}$,
Now eliminating the like terms we get,
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{{\left( { - 1} \right)\left( {n + 1} \right)x}}{n}$,
Now again simplifying for applying limit we get,
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( { - 1} \right)x\left( {\dfrac{n}{n} + \dfrac{1}{n}} \right)$,
Now applying limit we get,
$\Rightarrow \left( { - 1} \right)x\left( {1 + \dfrac{1}{\infty }} \right)$,
We know that$\dfrac{1}{\infty } = 0$, we get,
$\Rightarrow \left( { - 1} \right)x = - x$
By the ratio test,$\left| x \right| < 1$ for the series to be convergent. Therefore, the radius of convergence is 1.
The Malclaurin series is $\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n + 1}}\left( n \right){x^{n - 1}}}$ and the radius of convergence is 1.
Final Answer:

The Maclaurin series for $f\left( x \right) = \dfrac{1}{{{{\left( {1 + x} \right)}^2}}}$ are $\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n + 1}}\left( n \right){x^{n - 1}}}$ and radius of convergence for $f\left( x \right) = \dfrac{1}{{{{\left( {1 + x} \right)}^2}}}$ is 1.

Note:
A Maclaurin series is a power series that allows one to calculate an approximation of a function $f\left( x \right)$ for input values close to zero, given that one knows the values of the successive derivatives of the function at zero. In many practical applications, it is equivalent to the function it represents
A Maclaurin series can be used to approximate a function, find the antiderivative of a complicated function, or compute an otherwise incomputable sum. Partial sums of a Maclaurin series provide polynomial approximations for the function.
A Maclaurin series is a special case of a Taylor series, obtained by setting $x = 0$.
The radius of convergence is half the length of the interval; it is also the radius of the circle in the complex plane within which the series converges.
Convergence may be determined by a variety of methods, but the ratio test tends to provide an immediate value $r$ for the radius of convergence. The interval of convergence may then be determined by testing the value of the series at the endpoints $- r$ and $r$.