# Taylor Series of ln(1+x) – Formula and Derivation

-

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).

Hey 👋

I have always been passionate about statistics and mathematics education.

I created this website to explain mathematical and statistical concepts in the simplest possible manner.

If you've found value from reading my content, feel free to support me in even the smallest way you can.