Is it possible to solve a recurrence with constant ${a,b,c}$?
${T(n)=aT(n/2)+bn^c}$ , ${T(1)=1}$
I try to substitute all values of ${a,b,c}$ be ${1}$. But, I can't change back the form with constant.
Could anyone help me?
How to solve ${T(n)=aT(n/2)+bn^c}$, ${T(1)=1}$?
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$\begingroup$
recurrence-relations
generating-functions
closed-form
recursion
1 Answers
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Hint:
Let $n=2^k\Rightarrow\frac{n}{2}=2^{k-1}$
We can transform $${T_n=aT_\frac{n}{2}+bn^c} \ \text{ with }\ T_1=1$$ to $${t_k=at_{k-1}+b(2^k)^c} \ \text{ with }\ t_0=1$$
Then proceed with your usual methods. If stuck somewhere, check this out.