> [!NOTE] Definition (Real Natural Logarithm as Inverse of Real Exponential Function) > Let $x>0.$ The natural logarithm of $x$ is defined as $\ln (x)=y : \exp(y)=x$where $\exp$ denotes the [[Real Exponential Function|real exponential function]]. **Note**: [[Inverse of Real Exponential Function Exists, is Strictly Increasing and Continuous]], $\log x$ exists and is strictly increasing and continuous. # Properties By [[Real Natural Logarithm is Strictly Increasing]], for all $x>y>0,$ $\ln(x)>\ln(y).$ By [[Real Natural Logarithm of Product]], for all $u,v>0,$ $\log(uv)=\log(u)+\log(v).$ By [[Derivative of Real Natural Logarithm Function]], $\frac{d}{dx} \log (x)=\frac{1}{x}.$ > [!NOTE] Theorem (Log series) > For $0<x\leq 2$ $\log x = (x-1) - \frac{(x-1)^{2}}{2} + \frac{(x+1)^{3}}{3} - \dots$ > Equivalently for $-1\leq x<1$ $-\log(1-t) = t + \frac{t^{2}}{2} + \frac{t^{3}}{3} + \dots$ >*Proof*. Let $f$ be the function $f: x\mapsto \log x$. Note that the $n$th derivative of $f$ is $f^{(n)}(x) = (-1)^{n-1} (n-1)! \, x^{-n}$ so $f^{(n)} (1)=(-1)^{n-1}(n-1)!.$ > >The $n$th [[Taylor's Theorem With Lagrange Remainder for Real Function#^c01cd3|Cauchy remainder Taylor formula]] centred at $1$ is therefore $\begin{align} \log x =\; & 0 + (x-1) - \frac{(x-1)^{2}}{2} +\dots + (-1)^{n-2} \frac{(x-1)^{n-1}}{n-1} \\ &+(-1)^{n-1} t^{-n} \frac{(x-1)^{n}}{n} \\ = \; & (x-1) - \frac{(x-1)^{2}}{2} +\dots + (-1)^{n-2} \frac{(x-1)^{n-1}}{n-1} \\ &+ (-1)^{n-1} \frac{(x-t)^{n-1}(x-1)}{t^{n}} \end{align}$for some $t$ between $1$ and $x$. > >Now $\frac{(x-t)^{n-1}(x-1)}{t^n}=\frac{x-1}t\left(\frac{x-t}t\right)^{n-1}=\frac{x-1}t(-1)^{n-1}\left(1-\frac xt\right)^{n-1}$where $0<x<t<1.$ Since $x<t<1$ we have $0<1-x/t<1-x$ and so the error tends to zero as $n\to \infty$ as fast as the exponential $(1-x)^{n-1}.$ Therefore the infinite series converges for $0<x<1.$ # Applications - [[Discrete Logarithm]]. - [[Euler-Mascheroni Constant]].