How to find the Maclaurin series and the radius of convergence for #f(x)=1/(1+x)^2#?

Question

Isaiah Allen · Accepted Answer

The Maclaurin series is given by #sum_(n= 1)^oo (-1)^(n+ 1)(n)x^(n - 1) # and the radius of convergence is #1#

Recall that the McLaurin series is given by

#f(0) + (f'(0)x)/(1!) + (f''(0) x^2)/(2!) +... + (f^n(0) x^n)/(n!)#

Let's start by finding the value of #f(0)# on the first few derivatives.

#f(0) = 1/(1 + 0)^2 = 1#

#f'(x) = (-2(x + 1))/(x + 1)^4 = -2/(x + 1)^3# #f'(0) = -2/(0 +1)^3 = -2#

#f''(x) = 6(x + 1)^2/(x + 1)^6 = 6/(x + 1)^4# #f''(0) = 6#

Now let's recall the first few factorials.

#1! = 1# #2! = 2# #3! = 6# #...#

We can see the similarity here. Our first few terms are therefore

#1 - 2x + (6x^2)/2 - (24x^3)/6# #=1 - 2x + 3x^2 - 4x^3#

Which can be written as #sum_(n= 1)^oo (-1)^(n+ 1)(n)x^(n - 1) #

The radius of converge is given by the ratio test.

#=lim_(n -> oo) ((-1)^(n + 1+ 1)(n +1)x^(n + 1- 1))/((-1)^(n +1)n(x^(n - 1))#

#=lim_(n -> oo) ((-1)^(n + 2)(n + 1)x^n)/((-1)^(n + 1)(n)x^(n - 1))#

#=lim_(n-> oo) (-1)^( 1)(n + 1)/n x#

#=(-1)(1)x#

#=-x#

By the ratio test, #|x| < 1# for the series to be convergent. Therefore, the radius of convergence is #1#.

Hopefully this helps!

Samantha Adkison · Answer

undefined To find the Maclaurin series for $ f(x) = \frac{1}{(1 + x)^2} $, we can start by expressing it as a geometric series.

$$
\frac{1}{(1 + x)^2} = (1 + x)^{-2} = \sum_{n=0}^{\infty} \binom{-2}{n} x^n
$$

The Maclaurin series coefficients are given by the formula $ \binom{-2}{n} = \frac{(-2)(-2-1)(-2-2)\ldots(-2-n+1)}{n!} $.

Simplify the expression:

$$
\binom{-2}{n} = \frac{(-2)(-3)(-4)\ldots(-n)}{n!} = \frac{(-1)^n(2)(3)(4)\ldots(n)}{n!}
$$

This simplifies to:

$$
\binom{-2}{n} = (-1)^n \cdot \frac{n!}{(n)!} = (-1)^n \cdot n
$$

So, the Maclaurin series for $ f(x) = \frac{1}{(1 + x)^2} $ is:

$$
\sum_{n=0}^{\infty} (-1)^n \cdot n \cdot x^n
$$

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

$$
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|
$$

where $ a_n = (-1)^n \cdot n \cdot x^n $.

$$
\lim_{n \to \infty} \left| \frac{(-1)^{n+1} \cdot (n+1) \cdot x^{n+1}}{(-1)^n \cdot n \cdot x^n} \right|
$$

$$
= \lim_{n \to \infty} \left| \frac{(n+1) \cdot x}{n} \right|
$$

$$
= |x| \lim_{n \to \infty} \frac{n+1}{n}
$$

$$
= |x|
$$

Since the series converges for $ |x| < 1 $, the radius of convergence is $ R = 1 $.