differentiate the maclaurin series for 1/(1-x) twice to find the maclaurin series of 1/(1-x)^3. Thanks

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- April 26th 2010, 02:27 PMBrothamaclaurin series
differentiate the maclaurin series for 1/(1-x) twice to find the maclaurin series of 1/(1-x)^3. Thanks

- April 26th 2010, 04:34 PMDeadstar

If you differentiate twice you get

You just differentiate each term in the sum above and divide by 2. After a few terms you should see a pattern then just turn it into an infinite sum. - April 26th 2010, 06:05 PMBrotha
Thanks, but what would the maclaurin series be? Is it something like

sum_(n=0)^infinity1/2 n (n-1) x^(n-2). How do you get that? - April 26th 2010, 06:21 PMDeadstar
etc...

So differentiate both sides...

etc...

differentiate again...

etc...

This is a series of the form...

.

How did I get this? Just differentiate twice. You don't even need to all that stuff I did above. Just it's good to see it done so you know what's going on.

Differentiating twice gives you . I.e.

So summarizing...

=>