Taylor series expansion, f(x)=x/1-x at a=1?

f(x) is a function. so you can plug any number that will substitute as x. using your equation, substitue 2 for x. so if f(2)=2/1-2. doing the math you would get f(2)=2/-1 or -2. so to graph that, of your x=2 than your y would equal -2. so now substitute 1 for your x. f(1)=1/1-1. doing the math you would get 1/0. math law states that any fraction where the denominator (the lower number) is zero, it is undefined. you for your equation, you would have plot points on a graph for any number accept x=1. just remember, when you solve for x in a function you will get your y. hope this helps.

The problem with taking the series expansion of this is that every derivative it will be undefined. However, before learning these type of series expansions you should have been taught :

1 / (1 - x) =

∞

Σ.....xⁿ

n=0

1 + x + x² + x³ + x^4

The way we got this by seeing this is a geometric series where a = 1 and r =x.

a / (1 - r)

Since | r | = | x | < 1, it converges.

Now we're missing an x on top so all we do is bring it in :

x / (1 - x) =

∞

Σ.....x(ⁿ+¹)

n=0

Now let's expand :

x + x² + x³ + x^4 + x^5

at x=1, x/1-x is undefined!

freak u