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I want to solve the recurrence for Strassen's method (for multiplying square matrices) with domain transformation and get a closed form. The equation is given below:

$T(n)=b$, at $n=2$

$T(n)=7T(n/2)+an^2$, at $n>2$

I know that since I have to cancel out n/2, I should set $n=2^k$. I'll call that $G(k)$.

$G(k)=7G(k-1)+a(2^k)^2$

Now I'm not sure what to do. The $(2^k)^2$ term is quite confusing. It prevents telescoping or characteristic equations as far as I can tell. It doesn't seem to fit any of the simple forms for generating functions.

Whitesizzle
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    Is the $(2^k)^2$ what bothers you? $(2^k)^2=4^k$. The $7$ bothers me. But if we multiply by $7^{-k}$ and put $H(k)=7^{-k}G(k)$ we get $H(k)-H(k-1)=a\left(\frac{4}{7}\right)^k$. – Alamos Apr 30 '15 at 06:11
  • @Alamos The $4^k$ term bothers me because then the transformed recurrence is a polynomial in k, which is difficult to solve by hand. – Whitesizzle Apr 30 '15 at 16:52
  • This recurrence was also treated at the following MSE link. – Marko Riedel Apr 30 '15 at 19:55

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At each step, we replace $n$ with $\tfrac{n}{2}$ and scale each term by $7$ as follows: \begin{align*} T(n) &= 7T\left(\frac{n}{2}\right) + an^2 \\ 7T\left( \frac{n}{2} \right) &= 7^2T\left(\frac{n}{2^2}\right) + \frac{7}{4}an^2 \\ 7^2T\left( \frac{n}{2^2} \right) &= 7^3T\left(\frac{n}{2^3}\right) + \left(\frac{7}{4} \right)^2 an^2 \\ 7^3T\left( \frac{n}{2^3} \right) &= 7^4T\left(\frac{n}{2^4}\right) + \left(\frac{7}{4} \right)^3 an^2 \\ &~~\, \vdots \\ 7^{\log_2 n - 2}T(4) &= 7^{\log_2 n - 1}T(2) + \left(\frac{7}{4} \right)^{\log_2 n - 2} an^2 \\ 7^{\log_2 n - 1}T(2) &= 7^{\log_2 n - 1}b \\ \end{align*} Summing everything together, we observe that the terms telescope, yielding: \begin{align*} T(n) &= 7^{\log_2 n - 1}b + \left(1 + \tfrac{7}{4} + (\tfrac{7}{4})^2 + \cdots + (\tfrac{7}{4})^{\log_2 n - 2} \right)an^2 \\ &= 7^{\log_2 n - 1}b + \frac{(\tfrac{7}{4})^{\log_2 n - 1} - 1}{\tfrac{7}{4} - 1}an^2 \\ &= 7^{\log_2 n - 1}b + ((\tfrac{7}{4})^{\log_2 n - 1} - 1)\tfrac{3}{4}an^2 \\ &= \tfrac{1}{7}b7^{\log_2 n} + (\tfrac{4}{7}(\tfrac{7}{4})^{\log_2 n} - 1)\tfrac{3}{4}an^2 \\ &= \tfrac{1}{7}bn^{\log_2 7} + (\tfrac{4}{7}n^{\log_2 7/4} - 1)\tfrac{3}{4}an^2 \\ &= \tfrac{1}{7}bn^{\log_2 7} + \tfrac{3}{7}an^{2 + \log_2 7/4} - \tfrac{3}{4}an^2 \\ &\in \Theta(n^{\log_2 7}) \end{align*}

Adriano
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