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"Quadratic" Ability Formulas  (Read 645 times)
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Xifanie (Webmistress) [Posts: 4248]
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  • [April 20, 2017, 10:25:58 PM]
"Quadratic" Ability Formulas
« on: April 20, 2017, 10:25:58 PM »
I understand that most people here didn't take higher levels of math in high school, but it is really embarrassing to have seen this mistake become such a huge phenomenon.
I don't know who is the first person to have ever said that fist/truth/untruth/tiamat formulas were "quadratic" on GameFaqs, but because of this idiot, now just too many people use the word erroneously.

It doesn't take long to see how wrong it is to use the term quadratic in this case: https://en.wikipedia.org/wiki/Quadratic_formula
Just look at the charts; quadratic formulas increase, decrease and increase again or decrease, increase and decrease again. I've done this shit in high school.

FFT formulas are one way: they're exponential.

Just figure I'd toss that out there as someone who paid attention to "advanced" math classes in high school and graduated. :)

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    Timbo [Posts: 521]
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    • [April 21, 2017, 02:55:12 AM]
    Re: "Quadratic" Ability Formulas
    « Reply #1 on: April 21, 2017, 02:55:12 AM »
     Finally! I thought I was going crazy!
    Vanya [Posts: 3956]
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    • [April 21, 2017, 04:49:00 AM]
    Re: "Quadratic" Ability Formulas
    « Reply #2 on: April 21, 2017, 04:49:00 AM »
    We live in an age of misinformation and alternative facts.

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    Glain [Posts: 448]
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    • [April 21, 2017, 06:24:42 PM]
    Re: "Quadratic" Ability Formulas
    « Reply #3 on: April 21, 2017, 06:24:42 PM »
    Eh? Those formulas are indeed quadratic.

    A quadratic expression is a polynomial involving a squared term (and no higher exponent). That's actually a correct way to describe these formulas, since they do include an X squared term if you multiply them out. Using Truth as an example, each graph would be a parabola (and bottom out at MA=0) (they would bottom out at different points, generally the MA addend divided by -2) if you graphed the formulas out on, say, a graphing calculator (where MA would be the X-axis and would span negative to positive MA), but since you can't have negative MA, we don't normally worry about the negative MA part of the graph. Nonetheless, those formulas are quadratic (and not exponential - exponential means the exponent itself is the variable; something like 2^MA, which would start slow but grow to insane levels extremely fast).
    « Last Edit: April 27, 2017, 03:03:54 PM by Glain »
    Raijinili [Posts: 57]
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    • [April 27, 2017, 04:35:56 AM]
    Re: "Quadratic" Ability Formulas
    « Reply #4 on: April 27, 2017, 04:35:56 AM »
    Yeah, quadratic.

    For Wave Fist, the formula is (PA+2)/2 * PA (ignoring rounding). Multiplying it out gives 1/2*PA^2 + PA. Solving using THE quadratic formula gives roots at 0 and -2 PA, which are unreachable in-game.

    Exponential is much, much faster than any polynomial. An exponential growth rate would make every +1 PA turn into a constant multiplicative factor in damage.

    Not sure what to call Geomancy, which is (approximately) a second-order polynomial in two variables, but without any squared terms.
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    Rfh [Posts: 277]
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    • [May 01, 2017, 01:13:51 AM]
    Re: "Quadratic" Ability Formulas
    « Reply #5 on: May 01, 2017, 01:13:51 AM »

    Exponential is much, much faster than any polynomial.


    Well, a lot of polynomials have a finite intervale where they grow faster than a exponential function. For example 2*X^2 grow faster than e^x in (0.357 , 2.153), but in fact can be proven that there is a point where in any x value higher than this point a polynomial is lower than a exponential.

    But it doesn't happen with a convergent infinite degree polynomial, like Taylor series.

    For example:
    is higher than e^x in the interval (1 , ∞) 


    Not sure what to call Geomancy, which is (approximately) a second-order polynomial in two variables, but without any squared terms.

    Talking about how they growth, MA and PA is a function of Lv, but this function isn't linear (it's defined as a recurrence relation: http://finalfantasy.wikia.com/wiki/Stat_growth_(Tactics)) so we cannot talk about a quadratic formula in terms of Lv
    « Last Edit: May 01, 2017, 06:49:37 PM by Rfh »
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    • [May 05, 2017, 10:40:55 PM]
    Re: "Quadratic" Ability Formulas
    « Reply #6 on: May 05, 2017, 10:40:55 PM »
    But it doesn't happen with a convergent infinite degree polynomial, like Taylor series.

    Polynomials by definition have finitely many terms. If it's "infinite degree", it's a convergent power series. In fact, e^x is described by a convergent power series.

    Quote
    it's defined as a recurrence relation: http://finalfantasy.wikia.com/wiki/Stat_growth_(Tactics)) so we cannot talk about a quadratic formula in terms of Lv

    Being a recurrence relation doesn't prevent it from being polynomial.

    In fact, FFT stat growth is linear.
    Code: [Select]
    MA_lv = MA_1 * (lv + MAC) / (1 + MAC)
    Not just roughly linear, but actually linear even with rounding, because the bonus per level is constant as long as you don't change jobs.
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    Rfh [Posts: 277]
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    • [May 06, 2017, 08:15:33 AM]
    Re: "Quadratic" Ability Formulas
    « Reply #7 on: May 06, 2017, 08:15:33 AM »
    Polynomials by definition have finitely many terms. If it's "infinite degree", it's a convergent power series.  In fact, e^x is described by a convergent power series.



    Yes, and the power series are considered a generalization of polynomials.
    In fact, you can easily see that the power serie of my last post is just e^x^2  :lol: :lol:

    Being a recurrence relation doesn't prevent it from being polynomial.

    In fact, FFT stat growth is linear.
    Code: [Select]
    MA_lv = MA_1 * (lv + MAC) / (1 + MAC)
    Not just roughly linear, but actually linear even with rounding, because the bonus per level is constant as long as you don't change jobs.

    I saw the dependence of level and RawStat and I thought that the recurrence relation cannot be lineal but in fact it's a constant growth.

    Your formula it's clearly the explicit solution of the recurrence relation:



    Thank you for realizing my mistake.  :P


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