The Taylor series of ln(1+x) gives an expression for the function ln(1+x) expressed as a polynomial power series having infinitely many terms. The series is valid in the range x\in (-1,1] . In this article, we will give the formula for the Taylor series of ln(1+x) and explain how it can be obtained.
Formula for Taylor Series of ln(1+x):
The Taylor series for ln(1+x) about the point x=0 is given by the formula, ln(1+x) = x -\frac{x^2}{2} + \frac{x^3}{3}\ldots+ \frac{(-1)^{n+1}x^n}{n}+\ldots
The above series is valid only if x lies in the interval (-1,1] . Notice that the above series is also valid at x=1. On the other hand, it is not valid at the point x=-1, since the logarithm of zero is not defined.
We now explain how we can obtain the above formula via repeated differentiation. The below argument is not a rigorous proof but serves well as a heuristic argument.
We begin by assuming that the logarithmic function can be expanded in the form of an infinite polynomial as shown below. ln(1+x) = a_0 + a_1x + a_2x^2\ldots+ a_nx^n+\ldots
We need to determine the coefficients a_i of the above “infinite degree polynomial”.
Putting x=0 on both sides in the above series we get that a_0=ln(1) = 0
Let us assume that the above series can be differentiated term by term. Differentiating once on both sides we get, \frac{1}{1+x} = a_1 + 2a_2x\ldots+ na_nx^{n-1}+\ldots
Once again substituting x=0 on both sides, the higher order terms vanish and we get that a_1=1/(1+0) = 1
Differentiating again we get that, \frac{-1}{(1+x)^2} = 2a_2 + 6a_3x^{2}+\ldots. Once again substituting x=0 on both sides, the higher order terms vanish and we get that a_2=-1/2.
Continuing so on in this way one obtains the required Taylor Series about the point x=0.
Alternative Explanation:
Alternatively one can also obtain the series by term-by-term integration of the Taylor expansion of 1/(1+x). We know that for x\in (-1,1) we have the following formula from geometric series, \frac{1}{1+x} = 1 -x + x^2+\ldots+ (-1)^{n+1}x^n+\ldots
This series is uniformly convergent in the interval (-1,1) and hence we can integrate term by term to obtain that, \int_{0}^{x} \frac{1}{1+x} = \int_{0}^{x} 1 - \int_{0}^{x} x + \int_{0}^{x} x^2+\ldots+ \int_{0}^{x}(-1)^{n+1}x^n+\ldots ln(1+x) = x -\frac{x^2}{2} + \frac{x^3}{3}\ldots+ \frac{(-1)^{n+1}x^n}{n}+\ldots
This proves the formula for the Taylor expansion for the function ln(1+x).
