[Math] ELH: Wealth by level!

[CRGreathouse]

Does anyone have thoughts about extending the ELH's "wealth by level" chart past 40th?

I've come up with a formula that extends forever, but wanted to know if there are any good thoughts on it - I'm not particularly happy about my formula (the progression's generally fine, but the formula is more complicated than I'd prefer).

 

[Crothian]

from 39th to 40th I think is difference of like 1,300,000 So, what I did was just increase that by 100,000 each level. So from 40 to 41st the difference is 1.4 million. It was nice and simple which is what I was going for.

 

[Buddha the DM]

I have created a table for wealth & exp in MS Excel which I formatted for web viewing, and posted to my website. The address for the table is http://harineldermage.tripod.com/d20/Levels1-200.html

 

[Thanee]

I'd use this pretty linear progression:

Level 40: 13,600,000gp
Level 41: +1,500,000gp
Level 42: +1,500,000gp
Level 43: +1,700,000gp
Level 44: +1,700,000gp
Level 45: +1,900,000gp
Level 46: +1,900,000gp
... ... ...

At every odd level after 40, increase the gold added by an additional 200,000gp, starting at 1,300,000gp (the gold added for level 39 and 40 respectively). Every even level adds the same as the odd level before.

Or as a recursive formula for level 41+:

f(40) := 13,600,000gp
f := f(n-1) + 1,300,000gp + [ROUND_UP{(n - 40) / 2} * 200,000gp]; n >= 41

Or like this:

f(39) := 12,300,000gp
f(40) := 13,600,000gp
f := f(n-1) + [f(n-1) - f(n-2)] + [(n MOD 2) * 200,000gp]; n >= 41

Bye
Thanee

 

[Thanee]

This will yield the same results as Buddha the DM has on his table, BTW.

Bye
Thanee

 

[CRGreathouse]

At every odd level after 40, increase the gold added by an additional 200,000gp, starting at 1,300,000gp (the gold added for level 39 and 40 respectively). Every even level adds the same as the odd level before.

Wow, that's pretty flat. I guess that's what I should have expected, though; it really drops off after 20th.

((x^2-33x+290)/40 is a good approximation of this sequence (it lacks the round-off error, so it looks slightly different, but it will always be close to your pattern).

 

[Thanee]

Yeah, I really don't see how the available gold should increase exponentially.

But why do you use such an imprecise formula?
Should be possible to come up with one, that yields exactly those numbers and is not recursive.

I had thought of one, however, that only works for the even levels, didn't put enough thought into it to make it work for odd levels as well... here it is:

f := 13,600,000gp + [(n - 40) * 1,300,000gp] + [(m/2) * (m+1) * 400,000gp]; m = (n - 40) / 2; n >= 41; n is even

The following should work for odd levels, now you only need to make it into one that works for both!

f := 13,600,000gp + [(n - 40) * 1,300,000gp] + [(m/2) * (m+1) * 400,000gp] - [m * 200,000gp]; m = (n - 39) / 2; n >= 41; n is odd

Bye
Thanee

 

[CRGreathouse]

Yeah, I really don't see how the available gold should increase exponentially.

But why do you use such an imprecise formula?
Should be possible to come up with one, that yields exactly those numbers and is not recursive.

Well, no on both counts.

Its problem isn't that it's imprecise - it's that it's more precise than the standard formula. I knew that it could be accomplished with 2 formulae, but why should I complicate the system to get less precise results?

Second, it's not particularly possible to combine the two into a single formula. Rather, it might be possible (using absolute values and other unusual tricks), but the result would be harder to use than the original two formulae.

 

[Thanee]

With precise I meant exactly the numbers in the virtual table.
I suppose, you know that, tho!

Bye
Thanee

 

[CRGreathouse]

Here's a general formula:

f = 50,000 n ^ 2 - 2,550,000 n + 37,600,000 - 100,000 [[n / 2]]

where [[x]] is the greatest integer less than or equal to x.