# How do you do the Taylor series expansion for #f(x)=x/(1-x)# at a=1?

You cannot because

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To find the Taylor series expansion of ( f(x) = \frac{x}{1-x} ) at ( a = 1 ), we first need to find the derivatives of ( f(x) ) at ( a = 1 ). The general formula for the nth derivative of ( f(x) ) is ( f^{(n)}(x) = \frac{n!}{(1-x)^{n+1}} ).

Substitute ( a = 1 ) into the formula to find the nth derivative at ( a = 1 ):

[ f^{(n)}(1) = \frac{n!}{(1-1)^{n+1}} = n! ]

Now, we can use the Taylor series formula to write the Taylor series expansion of ( f(x) ) at ( a = 1 ):

[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(1)}{n!}(x-1)^n ]

Substitute the nth derivative at ( a = 1 ) into the formula:

[ f(x) = \sum_{n=0}^{\infty} \frac{n!}{n!}(x-1)^n ]

Simplify the expression:

[ f(x) = \sum_{n=0}^{\infty} (x-1)^n ]

This is the Taylor series expansion of ( f(x) = \frac{x}{1-x} ) at ( a = 1 ).

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