Is O(c^(k+1)) the same as O(c^k)?

Great question! The answer really depends on whether you see k as a variable or a constant. If k is a constant, then O(c^(k+1)) and O(c^k) are equivalent because multiplying by a constant doesn’t change the Big-O classification. However, if k varies, as you suggested, then they definitely show different growth rates.

To put it simply: when dealing with asymptotic notation, O(c^(k+1)) does not equal O(c^k) if both c and k are variables. If you take values—like c=2 and k=2—you can easily see O(2^(2+1)) is substantially larger than O(2^2). So the two notations can represent vastly different complexities.

Common convention classifies variables like 'c' or 'k' as constants. So if that's the case, people might view O(c^(k+1)) and O(c^k) as the same, because they leave us with expressions that boil down to constant growth rates. But if you treat k as a variable, then they are indeed different, especially if you consider specific values of k — just like O(n) and O(n²) are clearly not the same.

In Big-O analysis, you're defining how algorithms behave as inputs grow. If k is being considered a fixed number (like 1), you might think both are equal, but as k increases, O(c^(k+1)) grows much faster than O(c^k). This indicates they are not equal. Simply put, they don’t behave the same as k increases — think of them as two completely different growth rates.

If c and k are both variables, then O(c^(k+1)) is not the same as O(c^k). Think of it like the difference between O(n²) and O(n³) — they're not equivalent. You can actually prove this using the definition of Big-O. If they were the same, you'd have c^(k+1) in O(c^k), which leads to a contradiction because c can grow indefinitely. So, the two sets are definitely not equal.