# How do you find the maclaurin series expansion of #f(x)= x / (1-x^4)#?

Use the Maclaurin series for

#x/(1-x^4) = sum_(n=0)^oo x^(4n+1)#

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To find the Maclaurin series expansion of ( f(x) = \frac{x}{1-x^4} ), you can express the function as a geometric series and then expand it term by term.

The geometric series expansion for ( \frac{1}{1 - x} ) is ( 1 + x + x^2 + x^3 + \ldots ) with a common ratio of ( x ), valid for ( |x| < 1 ).

To adapt this for ( \frac{x}{1 - x^4} ), you need to rewrite ( x^4 ) as ( (x^2)^2 ), and then replace ( x ) with ( x^2 ) in the geometric series. This yields:

[ \frac{x}{1 - x^4} = x \cdot \frac{1}{1 - x^4} ]

[ = x \cdot \frac{1}{1 - (x^2)^2} ]

Now, replace ( x^2 ) with ( -x^2 ) in the geometric series:

[ = x \cdot \left(1 + (-x^2) + (-x^2)^2 + (-x^2)^3 + \ldots\right) ]

[ = x - x^3 + x^6 - x^9 + \ldots ]

This is the Maclaurin series expansion of ( f(x) = \frac{x}{1 - x^4} ).

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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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