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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: