# find the Taylor series at X=0 for the given finction, either by using the definition or by manipulating a known series F(x)=ln( (x+1)/(2x+1))

*print*Print*list*Cite

You need to find the Taylor's series of the given function about `x=0` such that:

`f(x) = f(0) + f'(0)*x + (f''(0))/(2!)*x^2 + (f^(3)(0))/(2!)*x^3+ ...`

Hence, you need to evaluate `f(0), f'(0),f''(0), f^(3)(0),...`

You need to substitute 0 for `x` in `f(x)` such that:

`f(0) = ln ((0+1)/(2*0+1)) => f(0) = ln (1/1) => f(0) = ln 1 = 0`

You need to evaluate `f'(x)` such that:

`f'(x) = 1/((x+1)/(2x+1))*((x+1)/(2x+1))'`

You need to use the quotient rule such that:

`((x+1)/(2x+1))' = ((x+1)'(2x+1) - (x+1)(2x+1)')/((2x+1)^2)`

`((x+1)/(2x+1))' = (2x + 1 - 2x - 2)/((2x+1)^2)`

`((x+1)/(2x+1))' = -1/((2x+1)^2)`

Hence, evaluating `f'(x)` yields:

`f'(x) = -(2x+1)/((2x+1)^2*(x+1)) => f'(x) = -1/((2x+1)(x+1))`

You need to evaluate `f'(0)` such that:

`f'(0)= -1/((2*0+1)(0+1)) = -1`

You need to evaluate `f''(x)` such that:

`f''(x) = ((-1)'*((2x+1)(x+1)) + ((2x+1)(x+1))')/(((2x+1)(x+1))^2)`

`f''(x) = (2x + 2 + 2x + 1)/(((2x+1)(x+1))^2)`

`f''(x) = (4x + 3)/(((2x+1)(x+1))^2)`

You need to evaluate `f''(0)` such that:

`f''(0) = (0+ 3)/(((0+1)(0+1))^2) = 3`

**Hence, evaluating the Taylor's series for the given function yields `f(x) = - x + 3/(2!)*x^2 + ......` **