Apologies for this simple question. Give a formula to predict the running time of a program for a problem of size N when doubling experiments have shown that the doubling factor is $2^b$ and the running time for problems of size N0 is T.
$$ \frac{T(2N)}{T(N)}=2^b$$
$$ T(N_0)=T $$
I am not sure how to proceed after that.
Edit:
I know that intuitively it would be $\bigl({\frac{N}{N_0}}\bigr)^bT$ but I am not sure how to write it out mathematically
Hint: For a given $N$ you have $\log_2(N/N_0)$ doublings of $N$. This corresponds to multiplying $T$ by $2^b$ that many times, giving $$ T(N) = T \cdot (2^b)^{\log_2(N/N_0)} $$ Now use power and logarithm rules to simplify the right-hand side of this.
If this is too hand-wavy for your context, you can prove that the above relation for $N$ of the form $2^k N_0$ by induction on $k$. When $N$ does not have precisely this form, the recurrence strictly speaking doesn't give you all the information you need, but if you have other reasons to expect that $T(\cdots)$ is increasing, knowing its value at $2^k N_0$ will be sufficient to speak about its asymptotic behavior.