Media The Skill Point Equation (nearly)

So @hppeng and I wanted exact skill point -> decimal conversion for reasons, and we did some math.


By pure luck, I guessed the base regression function correctly on my second try.
As we can see, we have a very close fit (R^2 = 1) meaning we have the (basically) real function (not the dumb quartic on the wiki) here.

However, there's still a problem: you see the values for a, b, c, and d?
They're floating points and lose precision from truncation (a lot of precision is probably lost in how desmos gave us a). If staff/anybody know the exact fractional/expression value for these numbers, could we please know them?

As proof floats cause problems:

Spoiler: Proof

source code:

Using the decimal function given by desmos results in 0.5024898(etc) for 68 skp, and the wiki (imported to the csv file) shows 0.503. This indicates round-off error somewhere.

(68 int is not necessarily the only mistake, it's just the first one I found)

 

Supplemental information posted below

 

Question for you guys, have you thought about trying to use a square root equation instead?

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Nvmd, math brain ain't working, that wouldn't work.

 

We are very very confident that the form of the equation is correct, even if the coefficients may be slightly off (The R^2 value suggests as much, and the equation gives the correct rounded values for every single skill point value except 68, and the "critical points" (41, 68, 94, 129) all round the correct way.). I'm not a stats boi but the fitting equation only has 4DOF and you're trying to fit 151 points and getting really close to each one.

It'd be funny if the form of the equation was actually something totally different (or there actually was no equation and its just a handmade scale).

 

Hmm, ok, well, I'll propose looking into the natural logarithm( f(x)=a*ln(x)+b ) as it does look like it could simplify it a little bit and the curve would fit just as well. I'd help a bit but my TI-83 doesn't give me the right equation and I haven't done had to use my calculus in months. Also, it's 2:15 where I'm at and need to get to bed.

 

it's really quite simple

 

Obligatory I don’t understand math comment, please keep comments under me about the actual post thank you

 

Yes, log will work too.

We get a curve very close to the exponential with this regression:

Overlay onto the exponential:

The curves differ outside of the data by a little. Maybe I'll try using the log expression.

aln(cx+d)+g had an R^2 of like .9996 and its path differed from the exponential more, so I'm trusting the one with f.

 

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