Recurrence relation complexity

Question

I just learned about recurrences and I just can't solve this problem. I have this recurrence relation:

$$ T(n) = \begin{cases} k\cdot T(\frac{n}{k}) & n > 0\\ 1 & n = 0\\ \end{cases} $$

where $k$ is a constant number.

I tried drawing a recurrence tree or replacing for lower $n$s but no success. I hope you can help me with an idea!

Do you really mean $T(nk)$? Is $k < 1$? – Yuval Filmus Jun 24 '15 at 18:02 — Yuval Filmus, Jun 24 '15 at 18:02
Sorry, i edited it wrong. Now is correct – Ruben P Jun 24 '15 at 18:27 — Ruben P, Jun 24 '15 at 18:27

score 2 · Accepted Answer · answered Jun 24 '15 at 18:29

Suppose that $n$ is a power of $k$, say $n = k^t$. Then $$ T(k^t) = kT(k^{t-1}) = k^2T(k^{t-2}) = \cdots = k^tT(1) = k^t, $$ assuming a base case of $T(1) = 1$. So for powers of $k$, we have $T(n) = n$. You can also prove that by induction: if $T(n/k) = n/k$ then $T(n) = kT(n/k) = k(n/k) = n$.

Recurrence relation complexity

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