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How do you find the maclaurin series expansion of #(x/(1+x^3))#?

Answer 1

Christopher Allers

Use the Maclaurin series for #1/(1-t)# and substitution to find:

#x/(1+x^3) = sum_(n=0)^oo (-1)^n x^(3n+1)#

The Maclaurin series for #1/(1-t)# is #sum_(n=0)^oo t^n#

since #(1-t) sum_(n=0)^oo t^n = sum_(n=0)^oo t^n - t sum_(n=0)^oo t^n = sum_(n=0)^oo t^n - sum_(n=1)^oo t^n = t^0 = 1#

Substitute #t = -x^3# to find:

#1/(1+x^3) = sum_(n=0)^oo (-x^3)^n = sum_(n=0)^oo (-1)^n x^(3n)#

Multiply by #x# to find:

#x/(1+x^3) = sum_(n=0)^oo (-1)^n x^(3n+1)#

This is a geometric series with common ratio #-x^3# so it converges if #abs(x) < 1# and has radius of convergence #1#.

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Answer 2

Emily Ammerman

To find the Maclaurin series expansion of ( \frac{x}{1+x^3} ), we first express it as a geometric series. Then, we find the Maclaurin series expansion of the geometric series.

The Maclaurin series expansion of ( \frac{x}{1+x^3} ) is given by:

[ \frac{x}{1+x^3} = x(1 - x^3 + x^6 - x^9 + \ldots) ]

Expanding this series gives:

[ x - x^4 + x^7 - x^{10} + \ldots ]

So, the Maclaurin series expansion of ( \frac{x}{1+x^3} ) is ( x - x^4 + x^7 - x^{10} + \ldots )

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Answer from HIX Tutor

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.