0

I’m trying to find a good approximation for $a-\log_{10}\left(10^a-1\right)$, with $a$ being a large positive real number.

I think the result should be in the order of magnitude of $10^{-a}$, but I can’t find a way to prove whether or not this is the case.

NoChance
  • 6,695
LSD_Sumus
  • 1
  • 2
    You could use $\ln(1-x)\approx -x$ for small $x$ – Andreas Lenz Apr 30 '23 at 12:16
  • 1
    See https://math.stackexchange.com/questions/977586/is-there-an-approximation-to-the-natural-log-function-at-large-values – NoChance Apr 30 '23 at 12:20
  • $f(x)-f(x-1)=f'(c)$, for $x-1<c<x$, and $f(t)=log_{10} x$, $f'(t)={1\over t .\ln(10)}$. For $x=10^a$, we get your result is approximately ${1\over 10^a .\ln(10)}$ – Thomas Apr 30 '23 at 15:05
  • @Thomas, thank you, that should do, do you happen to have any source for it? – LSD_Sumus Apr 30 '23 at 17:40

1 Answers1

1

Using the propertieds og logarithms, what you want to compute with high accuracy is $$\Delta(a)=a-\log_{10}\left(10^a-1\right)=-\frac{\log_e \left(1-10^{-a}\right)}{\log_e (10)}$$

Let $$1-10^{-a}=\frac {1+t}{1-t} \implies t=t(a)=\frac{1}{1-2\times 10^a}$$ and use $$\log \left(\frac{1+t}{1-t}\right)=2\sum_{n=0}^\infty \frac {t^{2n+1}}{2n+1}$$

Truncate the summation to some order and make the series as a $[2k+1,2k]$ Padé approximant $P_k$ such as $$P_2(a)=2\, t\,\, \frac{1-\frac{7 }{9}t^2+\frac{64 }{945}t^4 } { 1-\frac{10 }{9}t^2+\frac{5 }{21}t^4}$$

So, I shall compute for large numbers the absolute error $$\Phi(a)= \Delta(a)+\frac {P_2(a)}{\log_e(10)}$$

Computing a few numbers $$\left( \begin{array}{cc} a & \Phi(a) \\ 5 & 6.22\times 10^{-62} \\ 10 & 6.22\times 10^{-117} \\ 15 & 6.22\times 10^{-172} \\ 20 & 6.22\times 10^{-227} \\ 25 & 6.22\times 10^{-282} \\ 30 & 6.22\times 10^{-337} \\ 35 & 6.22\times 10^{-392} \\ 40 & 6.22\times 10^{-447} \\ 45 & 6.22\times 10^{-502} \\ 50 & 6.22\times 10^{-557} \\ \end{array} \right)$$

We can do much better using $P_{n>2}$

Claude Leibovici
  • 289,558