Ranter
Join devRant
Do all the things like
++ or  rants, post your own rants, comment on others' rants and build your customized dev avatar
Sign Up
Comments

Wisecrack335876d@Stuxnet
yes, but I love it.
I'm the Picasso of math. Fucking terrible at it. I want to be the *van gogh* of math.
Maybe if I cut off my ear and mail it to my love interest...
Edit: Is my math bad here or do you mean math in general? 
Wisecrack335876dApparently it becomes more accurate the larger the number.
It is accurate only for the unit digit, up to n=299.
At n = 446, it becomes accurate to two digits.
At n = 6975 it is three digits accurate and so on.
If anyone can find counter examples in these ranges where it is *less* accurate than the rest of the range, that would be cool. 
hitko146276dIt's way off for values below 1 (including negative numbers), other than that it's pretty close. But really, exponential function takes exactly as long to compute as its inverse, the logarithmic function  that given, it's faster to just use ln(n).

Wisecrack335876d@hitko
I'm so untrained at math I wasn't aware expotentials were the inverse of logarithms lol.
I just hate the idea of relying on a magic button. 
kraator68875dSimply plot the difference of your funktion and ln x:
On a linear scale (adjust x range at the end) :
https://wolframalpha.com/input/...
Using a logarithmic x axis :
https://wolframalpha.com/input/... 
Wisecrack335875d@kraator
You are a god.
I will now proceed to trade you gold in exchange for glass beads and wolfram alpha plots. 
Wisecrack335875d

Wisecrack335837d@kraator
Hey Kraator, it's probably a lot to ask, and I know I troll and shitpost a bit, in addition to being genuinely ignorant of a lot of things (even I can only tolerate so much selfembarrassment), but would you be willing to explain the difference for me about your two plots and what they mean?
Related Rants
Oh god, here comes another math post! I can feel it coming on, like werewolfism during the full moon.
I'm only passingly familiar with logarithms, so this, like everything I've stumbled on, has probably already been discovered, but
n/(1/((n^(1/n))1))
Is a pretty good approximation (within a couple percentage points, or three or more digits) of the natural logarithm for all the numbers I've checked it on.
For example if
n = 690841693
ln(n) = 20.35342125707679
while our estimate using the above formula comes out to:
n/(1/((n**(1/n))1)) = 20.353421612948146
Am I missing something obvious here, and if so, what?
Am I doing the idiot savant thing again, or am I just being an idiot again?
random
logarithms
algorithms
math