I want to prove that the time complexity of an algorithm is polylogarithmic in the scale of input.

The recurrence relation of this algorithm is $T(2n) \leq T(n-1) + T(n^a)$, where $a\in(0,1)$.

It seems that $T(n) \leq \log^{\beta}{n}$ for some $\beta$ depends on $a$. But I can't prove this inequality. How to solve this recurrence relation?

I want to prove that the time complexity of an algorithm is polylogarithmic in the scale of input.

The recurrence relation of this algorithm is $T(2n) \leq T(n) + T(n^a)$, where $a\in(0,1)$.

It seems that $T(n) \leq \log^{\beta}{n}$ for some $\beta$ depends on $a$. But I can't prove this inequality. How to solve this recurrence relation?

I just want to get an upper bound polylogarithmic in n.

I want to prove that the time complexity of an algorithm is polylogarithmic in the scale of input.

The recurrence relation of this algorithm is $T(2n) \leq T(n-1) + T(n^a)$.

It seems that $T(n) \leq \log^{\beta}{n}$, where $\beta$ depends on $a$. But I can't prove this inequality. How to solve this recurrence relation?

I want to prove that the time complexity of an algorithm is polylogarithmic in the scale of input.

The recurrence relation of this algorithm is $T(2n) \leq T(n-1) + T(n^a)$, where $a\in(0,1)$.

It seems that $T(n) \leq \log^{\beta}{n}$ for some $\beta$ depends on $a$. But I can't prove this inequality. How to solve this recurrence relation?