Mercator series

In mathematics, the Mercator series or Newton–Mercator series is the Taylor series for the natural logarithm:

$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$

In summation notation,

$\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} x^n.$

The series converges to the natural logarithm (shifted by 1) whenever $-1<x\le 1$ .

History

The series was discovered independently by Nicholas Mercator, Isaac Newton and Gregory Saint-Vincent. It was first published by Mercator, in his 1668 treatise Logarithmotechnia.

Derivation

The series can be obtained from Taylor's theorem, by inductively computing the n^th derivative of $\ln(x)$ at $x=1$ , starting with

$\frac{d}{dx}\ln(x)=\frac1{x}.$

Alternatively, one can start with the finite geometric series ($t\ne -1$)

$1-t+t^2-\cdots+(-t)^{n-1}=\frac{1-(-t)^n}{1+t}$

which gives

$\frac1{1+t}=1-t+t^2-\cdots+(-t)^{n-1}+\frac{(-t)^n}{1+t}.$

It follows that

$\int\limits_0^x \frac{dt}{1+t}=\int\limits_0^x \left(1-t+t^2-\cdots+(-t)^{n-1}+\frac{(-t)^n}{1+t}\right)\ dt$

and by termwise integration,

$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots+(-1)^{n-1}\frac{x^n}{n}+(-1)^n \int\limits_0^x \frac{t^n}{1+t}\ dt.$

If $-1<x\le 1$ , the remainder term tends to 0 as $n\to\infty$.

This expression may be integrated iteratively k more times to yield

$-xA_k(x)+B_k(x)\ln(1+x)=\sum_{n=1}^\infty (-1)^{n-1}\frac{x^{n+k}}{n(n+1)\cdots (n+k)},$

where

$A_k(x)=\frac1{k!}\sum_{m=0}^k{k\choose m}x^m\sum_{l=1}^{k-m}\frac{(-x)^{l-1}}{l}$

and

$B_k(x)=\frac1{k!}(1+x)^k$

are polynomials in x.^[1]

Special cases

Setting $x=1$ in the Mercator series yields the alternating harmonic series

$\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}=\ln(2).$

Complex series

The complex power series

$\sum_{n=1}^\infty \frac{z^n}{n}=z+\frac{z^2}{2}+\frac{z^3}{3}+\frac{z^4}{4}+\cdots$

is the Taylor series for $-\log(1-z)$ , where log denotes the principal branch of the complex logarithm. This series converges precisely for all complex number $|z|\le 1,z\ne 1$. In fact, as seen by the ratio test, it has radius of convergence equal to 1, therefore converges absolutely on every disk B(0, r) with radius r < 1. Moreover, it converges uniformly on every nibbled disk $\scriptstyle \overline{B(0,1)}\setminus B(1,\delta)$, with δ > 0. This follows at once from the algebraic identity:

$(1-z)\sum_{n=1}^m \frac{z^n}{n}=z-\sum_{n=2}^m \frac{z^n}{n(n-1)}-\frac{z^{m+1}}{m},$

observing that the right-hand side is uniformly convergent on the whole closed unit disk.

References

1. Medina, Luis A.; Moll, Victor H.; Rowland, Eric S. (2009). "Iterated primitives of logarithmic powers". arXiv:0911.1325.

• Weisstein, Eric W., "Mercator Series", MathWorld.
• Eriksson, Larsson & Wahde. Matematisk analys med tillämpningar, part 3. Gothenburg 2002. p. 10.
• Some Contemporaries of Descartes, Fermat, Pascal and Huygens from A Short Account of the History of Mathematics (4th edition, 1908) by W. W. Rouse Ball