Question

In Exercises 15-20, use the binomial series to find the Maclaurin series for the function. $$ f(x)=\frac{1}{(1+x)^{2}} $$

Short answer

The Maclaurin series for the function \( f(x)=\frac{1}{(1+x)^{2}} \) is \( -1 + 2x - 3x^2 + 4x^3 - ... \)

Step-by-step solution

Simplify the function

We rewrite the function as \(-\frac{d}{dx}[(1+x)^{-1}]\) so differentiation becomes easier in the next steps.

Apply Maclaurin Series for Binomial Expansion

Challenge is to express given function as a Maclaurin series. The Maclaurin series expansion for the function \( (1+x)^n \) where \( n \neq 0,-1,-2,-3,... \) is given by \( (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + .... \) In our case, n is -1. So, the Maclaurin series expansion of \( (1+x)^{-1} \) is \( 1 - x + x^2 - x^3 + x^4 - ... \)

Differentiating the series

Now, we need to differentiate the series with respect to x because in step 1 we re-wrote the function to include a differentiation. Differentiating \( 1 - x + x^2 - x^3 + x^4 - ... \) with respect to x, gives us \( 0 - 1 + 2x - 3x^2 + 4x^3 - ... \). We have to be aware of the alternating sign because of the negative exponent.

Writing out the final Maclaurin series

After the differentiation, we obtain \( -1 + 2x - 3x^2 + 4x^3 - ... \) which is the Maclaurin series for the initial function \( f(x) = \frac{1}{(1+x)^{2}} \)