$$\forall x\in\left(-1,1\right):\space\ln\left|1+x\right|=\int\frac{1}{1+x}\space dx=\int\sum_{n=0}^{\infty}\left(-x\right)^n\space dx=x-\frac{x^2}{2}+\frac{x^3}{3}\dots=\sum_{n=1}^{\infty}\left(-1\right)^{n-1}\cdot\frac{x^n}{n}$$

This is well known. I have two questions here. Firstly, I noticed that this is a convergent alternating series, and so the error of an approximation of degree $n$ is dominated by term $n+1$: $\xi_n(x)=\left|\frac{x^{n+1}}{n+1}\right|$ (I think!). However, when plotting this on Desmos, with $x\in\left(-1,1\right)$ of course, I noticed that the error predicted by this remainder function was not always an upper bound of the actual difference between the polynomial and the natural logarithm; the real error was greater than my "upper bound" sometimes, especially for low-degree approximations - the approximation at degree $1$ is visibly terrible, since it's a straight line, and the error is far greater than $\left|\frac{x^2}{2}\right|$ for $x$ near $-1$.

Secondly, I wondered if this was a useful approximation technique for $\ln\left|k+x\right|,\space\forall k>0$ - or maybe even $\forall k\in\mathbb{R}$? I'm not sure. Anyway, I attempted this in much the same way: $$\forall x\in\left(-\left|k\right|,\left|k\right|\right):\space\ln\left|k+x\right|=\int\frac{1}{k+x}\space dx=\int\frac{\frac{1}{k}}{1+\frac{x}{k}}\space dx=\int\sum_{n=0}^{\infty}\frac{1}{k}\cdot\left(-\frac{x}{k}\right)^n\space dx$$ $$\int\frac{1}{k}-\frac{x}{k^2}+\frac{x^2}{k^3}\dots\space dx=\frac{x}{k}-\frac{x^2}{2k^2}+\frac{x^3}{3k^3}\dots=\sum_{n=1}^{\infty}\left(-1\right)^{n-1}\cdot\frac{x^n}{n\cdot k^n}$$

The interval of convergence is $\left(-\left|k\right|,\left|k\right|\right)$ since $\left|-\frac{x}{k}\right|<1$, I think. Likewise, an error bound on this approximation fails miserably... and after plotting this on Desmos, I note that my approximation is terrible! It doesn't work at all :)

EDIT: This is despite my thought that $$\forall x,k>0,x<k:\lim_{n\to\infty}\xi_{n,k}(x)\space\left|\frac{x^{n+1}}{\left(n+1\right)k^{n+1}}\right|=0$$

Many thanks if anyone can clear up how to properly bound this!

  • $\begingroup$ Can you write down the bounds of your intervals? You may discover errors doing so. $\endgroup$ – mathcounterexamples.net Apr 28 at 6:40
  • $\begingroup$ @mathcounterexamples.net Are they not $x \in (0,|k|)$? $\endgroup$ – FShrike Apr 28 at 7:02
  • $\begingroup$ May I ask you to write the bounds yourself and compute properly the integral... rather than answering my question with another question? You'll discover your error by yourself then! $\endgroup$ – mathcounterexamples.net Apr 28 at 7:55
  • $\begingroup$ @mathcounterexamples.net after some empirical observations on Desmos I have deduced that my infinite sum is exactly equal not to $\ln\left|k+x\right|$ but instead to $\ln\left|\frac{k+x}{k}\right|$! $\endgroup$ – FShrike Apr 28 at 21:04
  • $\begingroup$ I believe that this is due to the algebraic rearrangement of $\frac{1}{k+x}$ to $\frac{1/k}{1+x/k}$ by multiplying both numerator and denominator by $\frac{1}{k}$, since $\frac{1}{k}$ is the difference between the failed approximation and the accurate one! However, a slight niggling doubt remains, since the two multiplications cancel, so the algebraic effect is nil - no? $\endgroup$ – FShrike Apr 28 at 21:11

First question

For $g(x)=x \in (-1,0)$, $$\sum_{n=1}^{\infty}\left(-1\right)^{n-1}\cdot\frac{x^n}{n}$$ is not an alternating series! $x^n$ itself has alternating signs and $g(x)$ tends to approach the harmonic series as $x \to 1$ which is diverging. This is coherent with what you found in Desmos.

Second question

If your goal is to evaluate $\ln(k+x)$, I would better use $\ln \frac{1}{x} = - \ln x$ to only deal with $x \gt 1$ and then $$\ln(a+x) = \ln a + \ln \left(1 + \frac{x}{a}\right).$$

  • $\begingroup$ Beat me to it. Excellent answer +1 $\endgroup$ – Alann Rosas Apr 27 at 19:15
  • $\begingroup$ The first response makes total sense and yes, I now realise that the series could never approximate the asymptote at $x=-1$. With regards to the second response, I do not understand the derivation there; shouldn't the second term be $\ln\left(1+\frac{x}{a}\right)$? Plus, my geometric series attempt for $\ln\left|k+x\right|$ was clearly wrong - but why? Do I just need a ridiculous number of terms to make it work, but if so, then why does the alternating bound fail even for positive $k, x$? $\endgroup$ – FShrike Apr 27 at 19:45
  • 1
    $\begingroup$ Thanks for spotting the typo. And yes, the approximation is very poor especially when $x$ is close to $k$ and even worse when close to $-k$. $\endgroup$ – mathcounterexamples.net Apr 27 at 19:53
  • $\begingroup$ @mathcounterexamples.net - poor due to an error in my workings, an error in my estimation of the error bound (because the alternating bound for my expansion actually goes to $\frac{1}{n}$ as $x\to k$) or some other reason? $\endgroup$ – FShrike Apr 27 at 20:00
  • $\begingroup$ I mean to say: my bound predicts a very good accuracy for high $n$, so why is the accuracy atrocious? $\endgroup$ – FShrike Apr 27 at 20:00

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