Enter a, b, and f(n)=nc. Watch the recursion tree grow level by level; each level's work ai·f(n/bi) is a bar. Whether the bars grow (leaves win), stay flat (tie), or shrink (root wins) reveals the case.
f(n)=n^c to the watershed n^(log_b a).
Each tree level i does total work a^i · f(n/b^i). Summing them is a geometric series with ratio r = a/b^c.
log factor, r→1 and the series
does not collapse — the basic theorem is silent. Use the extended Case 2 (adds one log) or
Akra–Bazzi (the integral form) instead.
a^i · (n/b^i)^c. r > 1 → bars grow → leaves win (Case 1).
r = 1 → bars flat → tie (Case 2, extra log factor). r < 1 → bars shrink → root wins (Case 3).
See junior.md, middle.md, and professional.md for the full derivation.