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

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.
  • #1
ivan77
17
0

Homework Statement


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


Homework Equations


maclaurin series


The Attempt at a Solution


part of the solution is to find the n derivatives of the function to setup the series
MY ANSWERS
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
TEXTBOOK ANSWERS
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.

Please help
 
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  • #2
ivan77 said:

Homework Statement


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


Homework Equations


maclaurin series


The Attempt at a Solution


part of the solution is to find the n derivatives of the function to setup the series
MY ANSWERS
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
TEXTBOOK ANSWERS
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.

Please help

The book is right. You are forgetting to use the chain rule.
 
  • #3
Dick said:
The book is right. You are forgetting to use the chain rule.

And in particular, that d/dx(1 - x) = -1
 
  • #4
I feel pretty silly. I very much appreciate the quick responses.
 

Related to Simple? maclaurin series (1-x)^-2

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:



This can also be written as:

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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