How can I do more accurate calculation Ln(1+x) in Verilog-A

[Jihwan_Park]

Hi,
I want to calculate ln(1+x)in verilog A. I know that the code is

y = ln(1+x);

But if the x value becomes very small(ex x=3.52e-18), the y value becomes zero. In MATLAB, I can calculate like

y=log1p(x);

and I want to calculate like that code. How can I calculate more accurately?

 

[wwfeldman]

you can examine the properties of natural logs and Taylor or McClaurin expansions

this may be a start

taylor series of $\ln(1+x)$?

Compute the taylor series of $\ln(1+x)$ I've first computed derivatives (up to the 4th) of ln(1+x) $f^{'}(x)$ = $\frac{1}{1+x}$ $f^{''}(x) = \frac{-1}{(1+x)^2}$ $f^{'''}(x) = \frac{2}{(1+x)^3}$ $...

math.stackexchange.com

 

[Jihwan_Park]

you can examine the properties of natural logs and Taylor or McClaurin expansions

this may be a start

taylor series of $\ln(1+x)$?

Compute the taylor series of $\ln(1+x)$ I've first computed derivatives (up to the 4th) of ln(1+x) $f^{'}(x)$ = $\frac{1}{1+x}$ $f^{''}(x) = \frac{-1}{(1+x)^2}$ $f^{'''}(x) = \frac{2}{(1+x)^3}$ $...

math.stackexchange.com

Thx it is very helpful for me!