# Halley’s formula

The following formula is due to the English scientist and mathematician Edmond Halley (1656 à 1742):

 $\displaystyle\ln{x}\;=\;\lim_{n\to\infty}(\sqrt[n]{x}-1)n$ (1)

Proof.  We change the $n$th root to power of $e$ and use the power series expansion of exponential function:

 $\displaystyle(\sqrt[n]{x}-1)n$ $\displaystyle\;=\;(e^{\frac{\ln{x}}{n}}-1)n$ $\displaystyle\;=\;\left(\sum_{m=0}^{\infty}\frac{(\ln{x}/n)^{m})}{m!}-1\right)\!n$ $\displaystyle\;=\;\sum_{m=1}^{\infty}\frac{(\ln{x}/n)^{m}n}{m!}$ $\displaystyle\;=\;\ln{x}+\frac{1}{n}\sum_{m=2}^{\infty}\frac{(\ln{x})^{m}}{m!n% ^{m-2}}$

The last converging series has a finite sum, and as  $n\to\infty$,  the asserted formula follows.

Note.  The formula (1) was known also by Leonhard Euler, who used it for defining the natural logarithm.  Using (1), one can easily prove the well-known laws of logarithm, e.g.

 $\displaystyle\ln{xy}$ $\displaystyle\;=\;\lim_{n\to\infty}(\sqrt[n]{x}\sqrt[n]{y}-1)n$ $\displaystyle\;=\;\lim_{n\to\infty}(\sqrt[n]{x}\sqrt[n]{y}-\sqrt[n]{y}+\sqrt[n% ]{y}-1)n$ $\displaystyle\;=\;\lim_{n\to\infty}y^{\frac{1}{n}}(\sqrt[n]{x}-1)n+\lim_{n\to% \infty}(\sqrt[n]{y}-1)n$ $\displaystyle\;=\;y^{0}\ln{x}+\ln{y}$ $\displaystyle\;=\;\ln{x}+\ln{y}.$

## References

• 1 Paul Loya: Amazing and Aesthetic Aspect of Analysis: On the incredible infinite. A course in undergraduate analysis, fall 2006. Available http://www.math.binghamton.edu/dennis/478.f07/EleAna.pdfhere.
Title Halley’s formula HalleysFormula 2013-03-22 19:34:39 2013-03-22 19:34:39 pahio (2872) pahio (2872) 8 pahio (2872) Result msc 40A05 ListOfCommonLimits