# natural logarithm

The natural logarithm    of a number is the logarithm in base of Euler’s number $e$. It can be defined as the map $\ln\colon\mathbb{R}_{+}\to\mathbb{R}$ satisfying

 $\ln(x)\colon=\int_{1}^{x}\frac{1}{t}dt.$ (1)

Figure 1 shows the graph of $\ln$. Figure 1: The graph of ln⁡(x).

Instead of $\ln$ many mathematicians write $\log$, physicists (and calculators) however consider $\log$ as the symbol for the logarithm in base 10. One can show that the function  defined in this way is the inverse     of the exponential function     . Indeed with equation (1) we have

 $\frac{d}{dx}\ln(e^{x})=e^{x}\cdot\frac{1}{e^{x}}=1,$

so there exists $C\in\mathbb{R}$ such that

 $\ln(e^{x})=x+C.$

Since $\ln(e^{0})=0$ we have $C=0$. One can also prove that the above integral  has the defining properties of a logarithm. For example if $x,y\in\mathbb{R}_{+}$ we have

 $\displaystyle\ln(xy)$ $\displaystyle=$ $\displaystyle\int_{1}^{x}\frac{1}{t}\mathit{dt}+\int_{x}^{xy}\frac{1}{t}% \mathit{dt}$ $\displaystyle=$ $\displaystyle\ln(x)+\int_{x}^{xy}\frac{1}{t}dt.$

Now applying the substitution law with $u\colon=\frac{t}{x}$ we have

 $\int_{x}^{xy}\frac{1}{t}\mathit{dt}=\int_{1}^{y}\frac{1}{u}\mathit{du}=\ln(y),$

so we have

 $\ln(xy)=\ln(x)+\ln(y).$

The natural logarithm can also be represented as a power series  around $1$. For $-1 we have

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

This series is divergent at $x=-1$ but for $x=1$ we have convergence due to Leibniz’s theorem  and obtain

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

In real analysis there is no reasonable way to extend the logarithm to negative numbers. In complex analysis the situation is a bit more complicated. Basically one can use the Euler relation to write a non-zero complex number   $z$ in the form $z=Re^{i\varphi}$ with $R,\varphi\in\mathbb{R}$. We could try to define the complex logarithm $\ln(z)$ to be $\ln(R)+i\varphi$. However $\varphi$ is unique only up to addition  of a multiple of $2\pi$. While at first glance this does not appear to be very problematic, it actually prevents one from setting up a continuous   logarithm on the complex plane  (without 0, where the logarithm should be infinite   ). Say for example that we let the imaginary part  of our logarithm take values from $-\pi$ to $\pi$. Then

 $\lim_{\varphi\searrow-\pi}\ln e^{i\varphi}=-i\pi$

and

 $\lim_{\varphi\nearrow\pi}\ln e^{i\varphi}=i\pi$

while $e^{i\varphi}\to-1$ for both limits. Therefore the logarithm we defined is not continuous at -1. The same argument  allows one to show that it is not continuous on the negative real numbers. In fact you can only define a continuous complex logarithm on a sliced plane, i.e. the complex plane with a half-line starting at 0 removed.

Title natural logarithm NaturalLogarithm 2013-03-22 12:28:28 2013-03-22 12:28:28 mathwizard (128) mathwizard (128) 13 mathwizard (128) Definition msc 33B10 MatrixLogarithm ComplexLogarithm