I have done the proof until the point when T(n) <= cn^log7

But when it comes to finding the value of constant c. I am getting stuck.

The given recurrence relation is

Since we already calculated the solution above which is . Inductive Step :

Now we have to prove that T(n) <= c n^log7 where c is a positive constant. If we consider that the solution holds good for n/2 then we can prove that it works for n also: T(n/2) <= c(n/2)^ log7 Substituting these values in the recurrence relation  :

T(n) <= 7c/(2)^log7 * (n)^log7 + n^2

 <= cn^log7 , since 7/(2)^log7  is a constant so it can be ignored and cn^log7 >> n^2 for large value of n
                                 
 <= cn^log7 assuming c is a constant >=1

. Finally to find constant c

(7/(2)^log7) * cn^log7 + n^2 <= cn^log7

So i am not able to find appropriate c for which the condition holds true.

I have done the proof until the point when $T(n) \leq cn^{\log7}$.

But when it comes to finding the value of constant $c$, I am getting stuck.

The given recurrence relation is $T(n) = 7T(n/2) + n^2$.

Since we already calculated the solution above which is $cn^{\log 7}$.

Inductive step:

Now we have to prove that $T(n) \leq c n^{\log7}$ where $c$ is a positive constant. If we consider that the solution holds good for $n/2$ then we can prove that it works for $n$ also: $$T(n/2) \leq c(n/2)^{\log7}.$$ Substituting these values in the recurrence relation:

$$ \begin{align*} T(n) &\leq 7c/(2)^{\log7} \times (n)^{\log7} + n^2 \\ &\leq cn^{\log7}, \text{ since $7/(2)^{\log7}$ is constant so can be ignored and $cn^{\log7} \gg n^2$ for large $n$} \\ &\leq cn^{\log7} \text{ assuming $c$ is a constant $\geq 1$.} \end{align*} $$

Finally to find constant $c$,

$$(7/(2)^{\log7}) \times cn^{\log7} + n^2 \leq cn^{\log7}. $$

I am not able to find appropriate $c$ for which the condition holds true.