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Find a series expansion for? : # (1-3x)^(2/3) #

Answer 1

Christopher Agresta

# (1-3x)^(2/3) = 1 - 2x - x^2 + 4/3x^3 + ... #

We seek an expansion of :

# (1-3x)^(2/3) #

The Binomial Series tells us that:

# (1+X)^n = 1+nX + (n(n-1))/(2!)X^2 (n(n-1)(n-2))/(3!)X^3 + ...#

And so for the given function we can expand using the Binomial Series as follows::

# f(x) = 1+(2/3)(-3x) + ((2/3)(-1/3))/(2!)(-3x)^2 + ((2/3)(-1/3)(-4/3))/(3!)(-3x)^3 + ... #

# \ \ \ \ \ \ = 1 - 2x - (2)/(2)x^2 + (8)/6x^3 + ... #

# \ \ \ \ \ \ = 1 - 2x - x^2 + 4/3x^3 + ... #

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Answer 2

Isaiah Allen

The series expansion for ( (1 - 3x)^{\frac{2}{3}} ) is:

[ 1 - 2x + \frac{10}{3}x^2 - \frac{20}{9}x^3 + \frac{140}{27}x^4 - \frac{560}{81}x^5 + \cdots ]

This expansion is valid for ( |x| < \frac{1}{3} ).

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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.