Calaculate a Ln without calculator

  • Thread starter Thread starter baby_1
  • Start date Start date
  • Tags Tags
    Calculator Ln
AI Thread Summary
The discussion focuses on calculating the natural logarithm (ln) of 0.163 without a calculator, leading to an approximate value of -1.9 for the equation involving z. Participants highlight that using approximations and properties of logarithms can yield quick results. The derived value for z is approximately 0.5, suggesting a significant relationship between the number 0.163 and the formula used. The conversation emphasizes the efficiency of manual calculations when familiar with logarithmic properties.
baby_1
Messages
159
Reaction score
16
Hello
Im solving a problem with calculator to find the zeta in this equation

gif.latex?e^{(\frac{-\xi\pi%20}{\sqrt{1-\xi^{2}}})}=.163.gif


but as i see some persons they solved this equation with different numbers(1.63 ) without calculator.now i want to know
how can fine zeta in this equation without calculator and fast?
 
Last edited by a moderator:
Physics news on Phys.org
ln(.163) = -2 + ln(e^2*0.163) ≈ -2 + ln(7*0.163) ≈ -2 + ln(1.1) ≈ -2 + 0.1 = -1.9
##-z\pi \approx -1.9 \sqrt{1-z^2}##
=> z>0
##z^2 \pi^2 \approx 3.6 - 3.6 z^2##
##z^2 (10+3.6) \approx 3.6##
##z^2 \approx \frac{3.6}{13.8} \approx \frac{4}{15} \approx \frac{2.7}{10} = 0.27##
##z \approx 0.5##
Solving speed ≈ typing speed
That is very close to the exact value and indicates that the .163 and the formula have some special meaning.

Edit: Fixed small error.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top