# How do you find a power series representation for #f(x)=1/(1+x)^3#?

Oh... I remember this one. Yeah, expect a long answer with this one.

I got:

So, what you would do is start from the one series you should have been taught, which was:

What you could do is figure out how to go from the original function to the function whose power series you DO know.

Therefore, what you have just done is:

- Integrated the original function
- Integrated the resultant function
- Multiplied by 2

Note that you will have to reverse these steps, so keep that in mind.

Next, we operate on the series to achieve the original function again.

To get the original function back, we have to undo what we did to the original function that allowed us to get to the series which we know the answer to, because operating on a power series mirrors operating on the function that is represented by said power series.

Therefore, we have to reverse the steps we performed earlier, but now on the power series rather than the function. To do this:

- Divide by 2
- Take the first derivative of the resultant function
- Take the first derivative of the resultant function to get back to the original function

- Divide by 2

And there you have it!

and Wolfram Alpha agrees with this answer.

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To find a power series representation for ( f(x) = \frac{1}{(1+x)^3} ), we can start by expressing ( \frac{1}{(1+x)^3} ) as a geometric series. We know that ( \frac{1}{1 - (-x)} = \sum_{n=0}^{\infty} (-x)^n ) for ( |x| < 1 ).

By differentiating both sides of this expression three times, we get:

( \frac{d^3}{dx^3} \left( \frac{1}{1 - (-x)} \right) = \frac{d^3}{dx^3} \left( \sum_{n=0}^{\infty} (-x)^n \right) )

This yields:

( \frac{3!}{(1 - (-x))^4} = \sum_{n=0}^{\infty} n(n-1)(n-2)(-x)^{n-3} )

Now, substituting ( x ) with ( -x ) and multiplying both sides by ( (-1)^3 ), we obtain:

( \frac{3!}{(1 + x)^4} = \sum_{n=0}^{\infty} n(n-1)(n-2)(-1)^{n-3}x^{n-3} )

Rearranging terms, we get:

( \frac{3!}{(1 + x)^4} = \sum_{n=0}^{\infty} \frac{n(n-1)(n-2)}{6} (-1)^{n-3}x^{n-3} )

Finally, multiplying both sides by ( \frac{1}{3!} ) (which is ( \frac{1}{6} )), we find the power series representation for ( f(x) = \frac{1}{(1+x)^3} ):

( f(x) = \sum_{n=0}^{\infty} \frac{n(n-1)(n-2)}{6} (-1)^{n-3}x^{n-3} )

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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