Timeline for Is there a mathematical formula to determine how much XP is needed per level?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Aug 25, 2021 at 23:03 | comment | added | F1000003 | When I tried it initially, I assumed it would be easy, and did something which looks similar to that Vandermonde matrix, running into much bigger compounding errors! So in the end I did some regression to fit it reasonably precisely on my 64-bit architecture without wanting to mess around with arbitrary precision - it fits much better. I've never learnt numerical analysis properly so I'm out of my comfort zone here - but given the size of the numbers involved, I think a more fundamental problem is that the exact Lagrange function solution requires more precision than my 64-bit words allow? | |
| Aug 25, 2021 at 22:10 | comment | added | Eddymage | I think it depends on how you computed it (for example, using the Vandermonde matrix leads to a ill conditioned problem), on rounding errors and on machine precision. | |
| Aug 25, 2021 at 22:00 | comment | added | F1000003 | Indeed that's what I did; (although presumably "of degree at most 19" is a necessary qualifier if you want to claim uniqueness). The classical result proves that such a formula exist; but I felt compelled to confess that the polynomial I've written doesn't quite pass through the required points. It isn't precise enough! | |
| Aug 25, 2021 at 21:52 | comment | added | Eddymage | Actually, it should give you the exact amount of XP per level when x=1,2,3,...,20. You just applied a classical result : it states that if you have \$n+1\$ couples of points \$(x_i,y_i)\$ then there exists a unique polynomial that passes (interpolates) the data. | |
| Aug 25, 2021 at 21:25 | history | edited | Someone_Evil♦ | CC BY-SA 4.0 |
MathJax to at least make it more readable
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| Aug 25, 2021 at 21:12 | history | answered | F1000003 | CC BY-SA 4.0 |