Prove or Disprove: For all real valued functions f, if f(n) is O(2¹), then f(2m) is 0(2m).

In step 2, I have provided answer with brief explanation......Answer: falseExplanation:Since given function is false, we have to disprove itThere is a counterexample to disprove given function..Let us consider a function f(n) = 2^n.We know that f(n) is O(2^n) [because it grows exponentially]However, if we take f(2m), we get f(2m) = 2^(2m)which is still an exponential growth function.So, f(2m) is O(2^(2m)), not O(2^m).In simple words, f(2m) grows at a faster rate (exponential) than 2^m (which is also exponential but with a smaller base),Therefore, it does not satisfy the condition that f(2m) is O(2^m).Conclusion:For all real valued functions f, if f(n) is O(2^n), then f(2m) is O(2^m)" has been disproven. A counterexample shows that for a function that is O(2^n), its double argument counterpart, f(2m), does not necessarily fall into the category of O(2^m). This counterexample demonstrates that the rate of growth in f(2m) can be significantly faster, making it not O(2^m). Therefore, the original statement does not hold true for all real-valued functions.

Answered: Prove or Disprove: For all real valued…

Prove or Disprove: For all real valued functions f, if f(n) is O(2¹), then f(2m) is O(2m)