Mazdak Farrokhzad edited a) Number of blits.tex  about 10 years ago

Commit id: 99fe0efd580eab58b89e0943caada47fa618793a

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For $n = 2^x$ we try the algorithm for incremental sizes of $x$ to find a pattern for how the algorithm behaves.  \begin{subequations}  \begin{split} \begin{align}  T(2^1) &= 5\\  T(2^2) &= 5 + 5 * 2^2\\  T(2^3) &= 5 + 5 * 2^2 + 5 * 2^4\\  T(2^4) &= 5 + 5 * 2^2 + 5 * 2^4 + 5 * 2^6\\  T(2^5) &= 5 + 5 * 2^2 + 5 * 2^4 + 5 * 2^6 + 5 * 2^8  \end{split} \end{align}  \end{subequations}  Given that: