> [!NOTE] Lemma ($\log$ [[Taylor's Theorem With Lagrange Remainder for Real Function|Taylor Series]] expansion with Langrange Remainder) > For $1\leq x\leq 2$ $\log x$ is given by the following [[Power Series|power series]]: $\log (x+1) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^4}{4} \dots$ ^837401 >*Proof*. Let $f$ be the function $f: x\mapsto \log x$. Then | Derivatives | Evaluated at 1 | | ---- | ---- | | $f'(x) = x^{-1}$ | $f(1)=0$ | | $f''(x) = -x^{-2}lt;br> | $f'(1)=0$ | | $f'''(x) = 2x^{-3}$ | $f''(1) = -1$ | | $f^{(n)}(x) = (-1)^{n-1} (n-1)! x^{-n}$ | $f^{(n)} (1)=(-1)^{n-1}(n-1)!$ | >The $n$th [[Taylor's theorem#^89aa6a|Lagrange remainder Taylor formula]] centred at $1$ is therefore $\begin{align} \log x &= 0 + (x-1) - \frac{(x-1)^{2}}{2} +\dots + (-1)^{n-2} \frac{(x-1)^{n-1}}{n-1} \\ &+(-1)^{n-1} t^{-n} \frac{(x-1)^{n}}{n} \\ &= (x-1) - \frac{(x-1)^{2}}{2} +\dots + (-1)^{n-2} \frac{(x-1)^{n-1}}{n-1} + (-1)^{n-1} \frac{1}{n} \left( \frac{x-1}{t} \right)^{n} \end{align}$for some $t$ between $1$ and $x$. > >If $1\leq x\leq 2$ then $0<x-1\leq 1\leq t$. Hence then error term is at most $\frac{1}{n}$ which tends to $0$ as $n\to \infty$. # Application - [[Real Natural Logarithm Function|Log Series]].