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Substation method fails to prove that $T(n)=\Theta(n^2) $ for the recursion $T(n)=4T(n/2) + n^2$, since you end up with $T(n) < cn^2 \leq cn^2 + n^2$.

I don't understand how to subtract off lower-order term to prove that substitution works.

Came up with: $T(n) \leq cn^2 - bn^2$

Assume it holds for $T(n/2) \leq c(n/2)^2 - b(n/2)^2$

$T(n) \leq 4(c(n/2)^2 - b(n/2)^2) + n^2 = cn^2- bn^2 + n^2 $

However, there is no way to solve $cn^2- bn^2 + n^2 \leq cn^2 - bn^2 $ for $b$

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    $\begingroup$ Are you sure it's in $\Theta(n^2)$? $\endgroup$ – saadtaame Sep 19 '12 at 12:29
  • $\begingroup$ very sure, it is taken right from the book $\endgroup$ – newprint Sep 19 '12 at 14:27
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    $\begingroup$ The solution is $\Theta(n^2 \log n)$. Try proving that by the substitution method. $\endgroup$ – saadtaame Sep 19 '12 at 17:23
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If you're using the same book I have (Introduction to Algorithms 3rd edition), then you mis-wrote the question. The relation in question should be: $T(n)=4T(n/2) + n.$

That said, substitution here won't work, as you know. Assuming $T(n) \le cn^2,$ you get:

$T(n)\le4T(n/2) + n$

$T(n)\le4(c\frac{n^2}{4}) + n$

$T(n)\le cn^2 + n$

Which doesn't imply $T(n) < cn^2.$ Instead, you can make the assumption: $T(n) \le cn^2 - bn.$

$T(n)\le4T(n/2) + n$

$T(n)\le4(c\frac{n^2}{4}-b\frac{n}{2}) + n$

$T(n)\le cn^2 + n(-2b + 1)$

Letting $b=1,$ you get an equation consistent with the initial asumption. Namely:

$T(n)\le cn^2 - n$

As a side note, the relation you initially wrote, $T(n)=4T(n/2) + n^2,$ can be solved using case 2 of Master's Theorem, and has an answer of $\Theta(n^2lg(n))$

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