Math Problem Statement

If f(n) = 3n^2 and g(n) = n^2, why in the world would we choose g(n) to upper bound f(n)? that doesn't intuitively make sense.

Solution

Great question! At first glance, it might not seem intuitive that g(n)=n2g(n) = n^2 could serve as an upper bound for f(n)=3n2f(n) = 3n^2, because f(n)f(n) grows faster. However, this comes down to how we define "upper bounds" in the context of asymptotic analysis and Big-O notation.

Big-O Notation and Upper Bounds:

When we say that one function "upper bounds" another in Big-O notation, we're not comparing them on a point-by-point basis, but rather focusing on their behavior as nn grows very large. Specifically, we're interested in finding a function that grows at least as fast as f(n)f(n) (or faster) in the long run, up to constant factors.

In this case:

  • f(n)=3n2f(n) = 3n^2 and
  • g(n)=n2g(n) = n^2

The key idea is that Big-O notation ignores constant coefficients. So, f(n)=3n2f(n) = 3n^2 can be expressed as O(n2)O(n^2), because for large nn, the factor of 3 becomes irrelevant when determining the growth rate.

Thus, g(n)=n2g(n) = n^2 does indeed provide an upper bound for f(n)=3n2f(n) = 3n^2, because in terms of growth rate, they both grow quadratically, and we don’t care about the constant multiplier in asymptotic analysis.

Why Choose g(n)g(n)?

  • Asymptotic Growth: Both f(n)f(n) and g(n)g(n) grow at the same rate, O(n2)O(n^2), which makes g(n)g(n) a valid upper bound.
  • Big-O Ignores Constants: The factor of 3 is irrelevant in Big-O notation, so O(n2)O(n^2) still captures the behavior of f(n)f(n).
  • Simplicity: We generally prefer to choose simpler upper bounds when possible. Since g(n)=n2g(n) = n^2 is simpler than 3n23n^2, it's a more convenient and valid choice.

Summary:

Choosing g(n)=n2g(n) = n^2 to upper bound f(n)=3n2f(n) = 3n^2 makes sense within the framework of Big-O notation, where constant factors are ignored, and only the dominant term for large nn matters. Both functions grow at the same quadratic rate.


Would you like more details on asymptotic analysis or Big-O notation? Here are 5 related questions to explore further:

  1. How does Big-O notation differ from Big-Theta and Big-Omega?
  2. What does it mean for a function to "dominate" another in terms of growth rate?
  3. Can two functions have the same Big-O but behave differently for small nn?
  4. Why is constant time complexity written as O(1)O(1)?
  5. How do you determine the Big-O of more complex functions, like polynomials or logarithms?

Tip: When using Big-O notation, always focus on the dominant term and ignore constants—they don't affect the asymptotic behavior!

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Math Problem Analysis

Mathematical Concepts

Asymptotic Analysis
Big-O Notation
Growth Rate

Formulas

f(n) = 3n^2
g(n) = n^2

Theorems

Big-O Upper Bound Theorem

Suitable Grade Level

Undergraduate Level (Computer Science or Mathematics)