# Master Theorem $T(n)=4T(n/8)+n^(3/8)$

My try was : $$f(n)= n^3/5=n^{0.6} g(n) = n^{\log_8}(4) =n^{0.667}$$ so $f(n)<g(n)$ So $f(n) = \Omega(n^{\log_8}(n) + \epsilon)$ but with regularity condition $4f(n/8) \le cn^{3/5}$ ,for $c$ constant $<1$ so$(4/3.48)n^{3/5} <= cn^{3/5}$ ...anomaly because $c<1$ not Case 3 Thanks

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f(n)= n^3/8 according to your title. that would make it case 1. –  chris Dec 8 '12 at 20:31