Armor equation question

ok, so the armor modifier equation for a caster assault is...

Actual Damage = Base Damage x 2 ^ ((Level*3 - Armor Rating ) /40)


My question is...if I am level 20 and I have AR60, wouldn't the equation result in zero?


AD = BD x 2 ^ ((60 - 60) / 40)

simplified...

AD = BD x 2 ^ (0 / 40)

and again...

AD = BD x 2 ^ 0


I can understand that if you have < AR60 then you would get a higher factor therefore the AD > BD and the opposite is apparent...but...

anything to the power of zero is zero...so multiply the base damage by zero and you get....zero

thoughts?

Anything to the power of zero is one.

meh, math ^ my memory > me

Sorry, but - maths might need a refresher for you

x^0 = 1 always.

So - when you are lvl 20, and facing 60 armour, your actual damage = your base damage

Ouch, don´t tell that a math professor. He will kill you! x^0 = 1 always? That is simply not right!

infinity^0 = ?

0^0 = ?

Infinity is not a number

-Diomedes

I know!

limes of exp(-x)^(1/x) for x = infinity => 0^0 = exp(-1)



limes of exp(x)^(1/x) for x = infinity => infinity^0 = exp(1)

0^0 is 1. At least, that is what is commonly accepted, since it solves so many problems - it is technically indeterminate, but 0^0=1 satisfies the binomial theorem and many other problems.

Edit: by satisfies, I mean that allowing x^0=1 for all x allows the binomial theorem to be true for x=0, y=0 and x=-y.

so many number my brans hurted now

I feel like I'm back in grade 10 math. My brain hurts!