Higher spell damage through math First let it be known that this doesn't apply to the cold tree, because its mastery contributes -res instead of a damage multiplier. We know that the formula for the total damage of a spell is: spell base damage * (1 + %bonus from synergy) * (1 + %bonus from mastery + %bonus from items) Now we break the individual bits down further: %bonus from synergy = synergy base + [additional] hard points in synergy * synergy bonus %bonus from mastery = mastery base + mastery bonus * ([additional] hard points in mastery + item points in mastery) So you're respecing and you have a pool of unallocated skill points to put into a certain spell. You've already put 20 points into the spell itself. Now the question is, with a minimum of one in mastery and each of the skill's prerequisites how many skill points should be put into synergy and how many into mastery if I want to maximize the damage multiplier for that particular spell? This question reveals another formula: [additional] hard points in synergy + [additional] hard points in mastery = total unallocated points To answer this question, first we need to algebratize it. x = (additional) hard points in synergy y = (additional) hard points in mastery For constants, I will use capital letters: S = synergy bonus M = mastery bonus I = item points in mastery B = mastery base = level 1 mastery bonus P = synergy base = the amount of synergy you're required to give the skill due to preqs T = %bonus from items (Temper bonus) U = unallocated points Now we restate the problem: (1+P+Sx)(1+B+T+M(y+I)) = Multiplier U = x + y y = -x + U (1+P+Sx)(1+B+T+M(-x+U+I)) As you learned in school, (a+b)(c+d) = ac + ad + bc + bd (1+P)(1+B+T) + (1+P)(M(-x+U+I)) + (Sx)(1+B+T) + (Sx)(M(-x+U+I)) And a(b + c) = ab + ac 1+B+T+P+PB+PT + -Mx+MU+MI-PMx+PMU+PMI + Sx+SBx+STx + Sx(M(-x+U+I)) The fourth term needs some additional factoring out: M(-x + U + I) = -Mx + MU + MI Sx(-Mx + MU + MI) = -SMxx + SMUx + SMIx Now we have: 1+B+T+P+PB+PT + -Mx+MU+MI-PMx+PMU+PMI + Sx+SBx+STx + -SM(x^2)+SMUx+SMIx This is a second degree polynomial. You could graph it if you so wished. To find the maximum value of a second degree polynomial with respect to x, we examine the following values: The earliest value in x's domain. 0 hard points beyond the prereqs in this case. The latest value in x's domain. U hard points beyond the prereqs in this case. The value where x "peaks." One quality of x's "peak" is that a line drawn tangent to that point will be flat. Its slope will be zero. Here is a picture of what I'm talking about. To find the slope of a tangent line of a function, we take that function's derivative. Here's the derivative, we want to know when it's equal to 0: -M + -P + S+SB+ST + -2SMx+SMU+SMI = 0 So we do algebra. -M-P+S+SB+ST+SMU+SMI = 2SMx (-M-P+S+SB+ST+SMU+SMI)/2SM = x That is the value of x that we are interested in, in terms of known constants. To be safe we'll want to examine both the floor and the ceiling of that value. Example coming later, I need a break from math right now.