# How do you find the maclaurin series expansion of #f(x)=(1-x)^-2#?

converging for

Start from the geometric series:

Note now that:

and inside the interval of convergence we can differentiate the series term by term, so:

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# f(x) = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + ...+ nx^(n-1) + ...#

We could alternatively derive a MacLaurin Series by using the Binomial Expansion: The binomial series tell us that:

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To find the Maclaurin series expansion of ( f(x) = (1 - x)^{-2} ), we can use the formula for the Maclaurin series expansion of ( (1 - x)^{-n} ), which is ( \sum_{k=0}^{\infty} \frac{(n+k-1)!}{(n-1)!k!} x^k ).

Substitute ( n = 2 ) into the formula:

( f(x) = (1 - x)^{-2} = \sum_{k=0}^{\infty} \frac{(2+k-1)!}{(2-1)!k!} x^k )

Simplify the expression:

( f(x) = \sum_{k=0}^{\infty} \frac{(k+1)!}{k!} x^k )

( f(x) = \sum_{k=0}^{\infty} (k+1) x^k )

So, the Maclaurin series expansion of ( f(x) = (1 - x)^{-2} ) is ( \sum_{k=0}^{\infty} (k+1) x^k ).

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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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