How do you use the binomial series to expand the function #f(x)=(1-x)^(2/3)# ?

Answer 1
This is a natural extension of raising a binomial to a whole number power: #(a+b)^n=sum_(k=0)^n (n/k)a^(n-k)b^k# (sorry about the inner bar) Where #(n/k)=(n!)/(k!(n-k)!)=(n(n-1)…(n-k+1))/(k!)#

So we can apply this to any exponent r even if r is an arbitrary real number.

#(a+b)^r=sum_(k=0)^oo (r(r-1)…(r-k+1))/(k!) a^(r-k)b^k#
Now put your info in, with #r=2/3, a=1, b=-x:#
#(1-x)^(2/3)=sum_(k=0)^oo (2/3(2/3-1)…(2/3-k+1))/(k!) 1^(2/3-k)(-x)^k#
#=(1)^(2/3)+(2/3)/1(1)^(-1/3)(-x)^1+((2/3)(2/3-1))/2(1)^(-4/3)(-x)^2+…#
#=1-2/3x-1/9x^2+…#

There's the start of the series; I dare you to compute the next two terms. Take the \dansmath challenge/!

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Answer 2

To expand the function ( f(x) = (1-x)^{\frac{2}{3}} ) using the binomial series, you can use the formula:

[ (1 + x)^{\alpha} = 1 + \alpha x + \frac{\alpha(\alpha - 1)}{2!}x^2 + \frac{\alpha(\alpha - 1)(\alpha - 2)}{3!}x^3 + \cdots ]

where ( \alpha = \frac{2}{3} ). Then, substitute ( -x ) for ( x ) in the formula to get the expansion for ( (1 - x)^{\frac{2}{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.

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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