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

ivan77
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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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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.
 
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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