MHB How can I develop ln(x) into a series for x >= 1 in fluid dynamics?

  • Thread starter Thread starter fluid_mechanics
  • Start date Start date
  • Tags Tags
    Series
fluid_mechanics
Messages
1
Reaction score
0
I need to develop $\mathrm{ln}(x)$ into series, where $x \geq 1$, and I don`t know how? In literature I only found series of $\mathrm{ln}(x)$, where:

1. $|x-1| \leq 1 \land x \neq 0$, $ \,\,\,\,\, \mathrm{ln}(x) = x - 1 - \dfrac{(x-1)^2}{2} + ...$ 2. $|x| \leq 1 \land x \neq -1$, $ \,\,\,\,\, \mathrm{ln}(x+1) = x - \dfrac{x^2}{2}+ ...$

My problem is problem in area of fluid dynamics, and $x$ is non-dimensional coordinate and it signifies radial coordinate of annular tube (it starts in the center of the tube). At the wall of inner tube $x=1$, and at the wall of outer tube it only can be larger (and values are not limited), because of that I need to fulfill a condition $x \geq 1$, for developing $\mathrm{ln}(x)$ into series.
 
Physics news on Phys.org
Write series for $$\ln (1+x)$$ and $$\ln (1-x)$$. Then subtract second from the first one and you'll obtain the series for $$\ln \left( \frac{1+x}{1-x} \right) $$, which gives you one possible series.
 
For original Zeta function, ζ(s)=1+1/2^s+1/3^s+1/4^s+... =1+e^(-slog2)+e^(-slog3)+e^(-slog4)+... , Re(s)>1 Riemann extended the Zeta function to the region where s≠1 using analytical extension. New Zeta function is in the form of contour integration, which appears simple but is actually more inconvenient to analyze than the original Zeta function. The original Zeta function already contains all the information about the distribution of prime numbers. So we only handle with original Zeta...
Back
Top