Mercator series

Polynomial approximation to logarithm with Template:Math, and Template:Math in the interval Template:Math.

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, $$\ln (1+x)=\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}}{n}{x}^n.$$

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

History

The series was discovered independently by Johannes Hudde (1656) and Isaac Newton (1665) but neither published the result. Nicholas Mercator also independently discovered it, and included values of the series for small values in his 1668 treatise Logarithmotechnia; the general series was included in John Wallis's 1668 review of the book in the Philosophical Transactions.

Derivation

The series can be obtained by computing the Taylor series of $\ln (x)$ at $x=1$: $$\ln (x)=(x-1)-\frac{(x-1{)}^2}{2}+\frac{(x-1{)}^3}{3}-\cdots ,$$ and substituting all $x$ with $x+1$. 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 $$\frac{1}{1+t}=1-t+t^2-\cdots +(-t{)}^{n-1}+\frac{(-t{)}^n}{1+t}.$$

It follows that $${\int }_0^x\frac{dt}{1+t}={\int }_0^x(1-t+t^2-\cdots +(-t{)}^{n-1}+\frac{(-t{)}^n}{1+t})dt$$ and by termwise integration, $$\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots +(-1{)}^{n-1}\frac{x^n}{n}+(-1{)}^n{\int }_0^x\frac{t^n}{1+t}dt.$$

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

This expression may be integrated iteratively Template:Mvar more times to yield $$-x{A}_k(x)+B_k(x)\ln (1+x)=\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n-1}\frac{x^{n+k}}{n(n+1)\cdots (n+k)},$$ where $$A_k(x)=\frac{1}{k!}\sum _{m=0}^k(\genfrac{}{}{0ex}{}{k}{m})x^m\sum _{l=1}^{k-m}\frac{(-x{)}^{l-1}}{l}$$ and $$B_k(x)=\frac{1}{k!}(1+x{)}^k$$ are polynomials in Template:Mvar.

Special cases

Setting $x=1$ in the Mercator series yields the alternating harmonic series $$\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{k+1}}{k}=\ln (2).$$

Complex series

The complex power series $$\sum _{n=1}^{\mathrm{\infty }}\frac{z^n}{n}=z+\frac{z^2}{2}+\frac{z^3}{3}+\frac{z^4}{4}+\cdots$$ 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 Template:Math with radius Template:Math. Moreover, it converges uniformly on every nibbled disk $\overline{B(0,1)}\setminus B(1,\delta )$, with Template:Math. 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.

See also

• John Craig
• Natural logarithm plus 1

References

• Template:Mathworld
• Anton von Braunmühl (1903) Vorlesungen über Geschichte der Trigonometrie, Seite 134, via Internet Archive
• 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