# Simple? maclaurin series (1-x)^-2

• ivan77
In summary, the Maclaurin series expansion for the function (1-x)^-2 is given by: n fn(x) 0 (1-x)^-2 (0th derivative)1 2(1-x)^-3 (1st derivative)2 6(1-x)^-4 (2nd derivative)3 24(1-x)^-5 (3rd derivative) The negative signs are due to the use of the chain rule in finding the derivatives.
ivan77

## Homework Statement

what is the maclaurin series expansion of the function (1-x)^-2

maclaurin series

## The Attempt at a Solution

part of the solution is to find the n derivatives of the function to setup the series
n fn(x)
0 (1-x)^-2 this should be the 0th derivative of the function
1 -2(1-x)^-3 this should be the first derivative of the function
2 6(1-x)^-4 second derivative
3 -24(1-x)^-5 third derivative
n fn(x)
0 (1-x)^-2 this should be the 0th derivative of the function
1 2(1-x)^-3 this should be the first derivative of the function
2 6(1-x)^-4 second derivative
3 24(1-x)^-5 third derivative
NOTICE the lack of negatives on the 1 and 3 derivative.

The thing that is confusing the heck out of me as well is that when I look for the first derivative of the function (1-x)^-2 on wolfram alpha it gives me -2(1-x)^-3, but when I look for the maclaurin series, the answer does not show the negative sign.

ivan77 said:

## Homework Statement

what is the maclaurin series expansion of the function (1-x)^-2

maclaurin series

## The Attempt at a Solution

part of the solution is to find the n derivatives of the function to setup the series
n fn(x)
0 (1-x)^-2 this should be the 0th derivative of the function
1 -2(1-x)^-3 this should be the first derivative of the function
2 6(1-x)^-4 second derivative
3 -24(1-x)^-5 third derivative
n fn(x)
0 (1-x)^-2 this should be the 0th derivative of the function
1 2(1-x)^-3 this should be the first derivative of the function
2 6(1-x)^-4 second derivative
3 24(1-x)^-5 third derivative
NOTICE the lack of negatives on the 1 and 3 derivative.

The thing that is confusing the heck out of me as well is that when I look for the first derivative of the function (1-x)^-2 on wolfram alpha it gives me -2(1-x)^-3, but when I look for the maclaurin series, the answer does not show the negative sign.

The book is right. You are forgetting to use the chain rule.

Dick said:
The book is right. You are forgetting to use the chain rule.

And in particular, that d/dx(1 - x) = -1

I feel pretty silly. I very much appreciate the quick responses.

## What is a Maclaurin series?

A Maclaurin series is a type of power series expansion that represents a function as an infinite sum of terms involving powers of a variable. It is named after the Scottish mathematician Colin Maclaurin.

## What is the formula for the Maclaurin series of (1-x)^-2?

The formula for the Maclaurin series of (1-x)^-2 is:

$\Large&space;f(x)=\sum_{n=0}^{\infty}(n+1)x^n$

This can also be written as:
$\Large&space;f(x)=\sum_{n=0}^{\infty}(-1)^n(n+1)x^n$

## What is the purpose of using a Maclaurin series?

The purpose of using a Maclaurin series is to approximate a complicated function with a simpler one. This allows for easier mathematical calculations and can also provide insights into the behavior of the function.

## How do you find the Maclaurin series of a function?

To find the Maclaurin series of a function, you can use the formula for the general term of the series and plug in the function's derivatives evaluated at x=0. Alternatively, you can use known Maclaurin series for basic functions and apply algebraic operations to obtain the series for more complex functions.

## What is the convergence of the Maclaurin series for (1-x)^-2?

The Maclaurin series for (1-x)^-2 has a radius of convergence of 1, meaning it converges for all values of x within a distance of 1 from the center (x=0). This can be determined by using the ratio test to check the convergence of the series.

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