Math Problem Statement

Prove that (log2 n)^3 is smaller than n when n goes very large. without using limit

Solution

To prove that (log2n)3(\log_2 n)^3 is smaller than nn when nn is very large without using limits, we can compare the growth rates of both functions through direct analysis and properties of logarithms and polynomials.

Step 1: Properties of Logarithms and Polynomials

We need to show that there exists a threshold n0n_0 such that for all n>n0n > n_0, (log2n)3<n(\log_2 n)^3 < n.

Step 2: Analyzing the Functions

Let's denote f(n)=(log2n)3f(n) = (\log_2 n)^3 and g(n)=ng(n) = n.

For large nn, the function g(n)=ng(n) = n grows exponentially faster than the function f(n)=(log2n)3f(n) = (\log_2 n)^3, which only grows polynomially with respect to log2n\log_2 n.

Step 3: Inequality Setup

Consider the inequality:

(log2n)3<n(\log_2 n)^3 < n

Taking the base-2 logarithm of both sides, we get:

log2((log2n)3)<log2(n)\log_2((\log_2 n)^3) < \log_2(n)

Using the power rule for logarithms, log2(ab)=blog2(a)\log_2(a^b) = b \log_2(a), we can simplify the left-hand side:

3log2(log2n)<log2n3 \log_2(\log_2 n) < \log_2 n

Step 4: Isolate the Logarithm Term

Now, we divide both sides by log2n\log_2 n:

3log2(log2n)log2n<13 \frac{\log_2(\log_2 n)}{\log_2 n} < 1

This inequality can be further simplified to:

log2(log2n)log2n<13\frac{\log_2(\log_2 n)}{\log_2 n} < \frac{1}{3}

Step 5: Bounding the Ratio

We need to show that for sufficiently large nn, the term log2(log2n)log2n\frac{\log_2(\log_2 n)}{\log_2 n} becomes very small. Observe that as nn grows, log2n\log_2 n grows much faster than log2(log2n)\log_2(\log_2 n).

Step 6: Direct Comparison for Large nn

To get a concrete understanding, consider the behavior for large nn:

  • For n=210=1024n = 2^{10} = 1024, log2n=10\log_2 n = 10, and log2(log2n)=log2(10)\log_2(\log_2 n) = \log_2(10).
  • For n=220n = 2^{20}, log2n=20\log_2 n = 20, and log2(log2n)=log2(20)\log_2(\log_2 n) = \log_2(20).

As nn increases, the ratio log2(log2n)log2n\frac{\log_2(\log_2 n)}{\log_2 n} decreases:

log2(20)204.32200.216\frac{\log_2(20)}{20} \approx \frac{4.32}{20} \approx 0.216

For larger nn, this ratio continues to decrease.

Step 7: Conclusion

Thus, for sufficiently large nn, log2(log2n)log2n<13\frac{\log_2(\log_2 n)}{\log_2 n} < \frac{1}{3}, which implies:

3log2(log2n)log2n<13 \frac{\log_2(\log_2 n)}{\log_2 n} < 1

Therefore,

(log2n)3<nfor sufficiently large  n(\log_2 n)^3 < n \quad \text{for sufficiently large} \; n

This completes the proof that (log2n)3(\log_2 n)^3 is smaller than nn for very large nn.

Would you like more details or have any questions? Here are five additional questions you might find interesting:

  1. Can you explain the differences between polynomial and exponential growth in more detail?
  2. How does the change of base formula for logarithms work in general?
  3. What are other examples of functions with different growth rates?
  4. How can we apply similar techniques to other logarithmic inequalities?
  5. What are the implications of understanding logarithmic versus polynomial growth in computer science?

Tip: When dealing with inequalities involving logarithms and polynomials, understanding their fundamental growth behaviors can often simplify proofs and comparisons.

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Math Problem Analysis

Mathematical Concepts

Logarithms
Polynomials
Inequalities

Formulas

Properties of logarithms
Power rule for logarithms

Theorems

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Suitable Grade Level

Advanced High School