. Find the Maclaurin series for f(x) = (1 + x) −3 using the definition. Find the radius of convergence.

Question

Nick Camelot · Accepted Answer

We are asked to find the Maclaurin series for

f (x) = (1 + x)^{- 3}

using the definition, and then find the radius of convergence.
The Maclaurin series for a function

f (x)

is given by the formula:

f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n},

where

f^{(n)} (0)

denotes the

n

th derivative of

f (x)

evaluated at

x = 0

.
First, we need to find the derivatives of

f (x)

up to the third order. We have:

f (x) = (1 + x)^{- 3},

so

f^{'} (x) = - 3 (1 + x)^{- 4},

f^{″} (x) = 12 (1 + x)^{- 5},

and

{f^{″}}^{'} (x) = - 60 (1 + x)^{- 6} .

Next, we evaluate these derivatives at

x = 0

:

f (0) = (1 + 0)^{- 3} = 1,

f^{'} (0) = - 3 (1 + 0)^{- 4} = - 3,

f^{″} (0) = 12 (1 + 0)^{- 5} = 12

and

{f^{″}}^{'} (0) = - 60 (1 + 0)^{- 6} = - 60 .

Now we can substitute these values into the formula for the Maclaurin series:

f (x) = f (0) + f^{'} (0) x + \frac{f^{″} (0)}{2!} x^{2} + \frac{{f^{″}}^{'} (0)}{3!} x^{3} + \dots

= 1 - 3 x + \frac{12}{2!} x^{2} - \frac{60}{3!} x^{3} + \dots

= 1 - 3 x + 2 x^{2} - 5 x^{3} + \dots .

Thus, the Maclaurin series for

f (x) = (1 + x)^{- 3}

is

1 - 3 x + 2 x^{2} - 5 x^{3} + \dots

.
To find the radius of convergence, we use the ratio test:

{lim}_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | = {lim}_{n \to \infty} | \frac{- 3 (n + 1)}{(n + 1) (n + 2)} | = 3 {lim}_{n \to \infty} \frac{1}{n + 2} = 0 .

Since the limit is zero, the series converges for all

x

values, so the radius of convergence is infinite.

. Find the Maclaurin series for f(x) =